Derivatives of Exponential & Logarithmic Functions: Videos & Practice Problems

To find the derivative of an exponential function, use the rule: if \( f(x) = b^x \), then \( f'(x) = b^x \ln(b) \), where \( b > 0 \) and \( b \neq 1 \). For the natural exponential function \( e^x \), the derivative is simply \( e^x \). The derivative of logarithmic functions follows the rule: if \( g(x) = \log_b(x) \), then \( g'(x) = \frac{1}{x \ln(b)} \). For the natural logarithm \( \ln(x) \), the derivative is \( \frac{1}{x} \), valid for \( x > 0 \).

Derivatives of General Exponential Functions

Understanding how to find the derivative of an exponential function is essential in calculus. The derivative of a function measures how the function's output changes as its input changes, and for exponential functions of the form \( b^x \), where \( b \) is a positive constant not equal to 1, there is a specific rule to follow.

The derivative of \( b^x \) can be derived using the limit definition of the derivative. This involves setting up the expression:

\[f'(x) = \lim_{h \to 0} \frac{b^{x+h} - b^x}{h}\]

By applying properties of exponents, this can be rewritten as:

\[f'(x) = \lim_{h \to 0} \frac{b^x \cdot b^h - b^x}{h} = b^x \cdot \lim_{h \to 0} \frac{b^h - 1}{h}\]

It turns out that the limit \( \lim_{h \to 0} \frac{b^h - 1}{h} \) is equal to \( \ln(b) \), where \( \ln \) denotes the natural logarithm. Therefore, the derivative of \( b^x \) is given by:

\[f'(x) = b^x \cdot \ln(b)\]

This formula highlights that the derivative of an exponential function is the function itself multiplied by the natural logarithm of its base. It is important to remember that the base \( b \) must be greater than 0 and not equal to 1 to ensure the function is well-defined.

For example, to find the derivative of \( f(x) = 6^x \), we apply the rule:

\[f'(x) = 6^x \cdot \ln(6)\]

In cases where the exponent is a more complex function, such as \( g(x) = 3^{x^2 + 4x} \), we must also apply the chain rule. The derivative in this case is calculated as follows:

\[g'(x) = 3^{x^2 + 4x} \cdot \ln(3) \cdot (2x + 4)\]

Here, \( 2x + 4 \) is the derivative of the inner function \( x^2 + 4x \). This demonstrates how to combine the rules for exponential functions and the chain rule effectively.

As you continue to practice finding derivatives, remember to apply these rules consistently, and you'll become proficient in handling various types of functions.

Find the derivative of the given function.
$f(x)=2^x-5^x$

• A. $x^2\ln2-x^5\ln5$
• B. $2^{x}\ln2-5^{x}\ln5$
• C. $2^{x}\ln x-5^{x}\lnx$
• D. $2\lnx-5\ln x$

Answer: (B)

Find the derivative of the given function.
$g(x)=x^4+4x+4^x$

• A. $12x$
• B. $4x^3+8$
• C. $4x^3+4x^{x-1}$
• D. $4x^3+4+4^{x}\ln4$

Answer: (D)

Find the derivative of the given function.
$y=(4x-3x^2+9)\cdot2^{5x}$

• A. $2^{5x}[5\ln 2(4x-3x^2+9)+(4-6x)]$
• B. $2^{5x}\cdot(-15x^2-14x+49)$
• C. $2^{5x}[(4-6x)+\ln(2)(4x-3x^2+9)]$
• D. $5\ln(2)(4-6x)\cdot2^{5x}$

Answer: (A)

Find the derivative of the given function.
$h\left(x\right)=4^{{(\sqrt{x}+3x)}^{\frac{5}{4}}}$

• A. $\frac54\left(\sqrt{x}+3x\right)\left(\frac{1}{2\sqrt{x}}+3\right)\cdot4\left(\sqrt{x}+3x\right)^{\frac54}$
• B. $\frac{5\ln\left(4\right)}{4}\left(\sqrt{x}+3x\right)\cdot4\left(\sqrt{x}+3x\right)^{\frac54}$
• C. $\frac54\cdot4\left(\sqrt{x}+3x\right)^{\frac54}\cdot\ln4\cdot\left(\sqrt{x}+3x\right)^{\frac14}\cdot\left(\frac{1}{2\sqrt{x}}+3\right)$
• D. $\left(\sqrt{x}+3x\right)^{\frac54}\cdot4\left(\sqrt{x}+3x\right)^{\frac14}$

Answer: (C)

Derivatives of General Exponential Functions Example 1

To find the derivative of the function \( f(\theta) = \sin(\theta \cdot 3^{\theta^2 + 4\theta}) \), we will utilize both the chain rule and the product rule due to the composition of functions involved.

First, we identify the outer function as \( \sin(u) \), where \( u = \theta \cdot 3^{\theta^2 + 4\theta} \). The derivative of \( \sin(u) \) is \( \cos(u) \), so we start with:

\( f'(\theta) = \cos(\theta \cdot 3^{\theta^2 + 4\theta}) \cdot u' \)

Next, we need to find \( u' \) using the product rule, which states that if \( u = v \cdot w \), then \( u' = v' \cdot w + v \cdot w' \). Here, let \( v = \theta \) and \( w = 3^{\theta^2 + 4\theta} \).

Calculating \( v' \) gives us:

\( v' = 1 \)

Now, for \( w = 3^{\theta^2 + 4\theta} \), we apply the chain rule. The derivative of \( 3^x \) is \( 3^x \ln(3) \), so:

\( w' = 3^{\theta^2 + 4\theta} \ln(3) \cdot (\theta^2 + 4\theta)' \)

Next, we find the derivative of the exponent \( \theta^2 + 4\theta \):

\( (\theta^2 + 4\theta)' = 2\theta + 4 \)

Thus, we have:

\( w' = 3^{\theta^2 + 4\theta} \ln(3) (2\theta + 4) \)

Now substituting back into the product rule for \( u' \):

\( u' = v' \cdot w + v \cdot w' = 1 \cdot 3^{\theta^2 + 4\theta} + \theta \cdot (3^{\theta^2 + 4\theta} \ln(3) (2\theta + 4)) \)

Factoring out \( 3^{\theta^2 + 4\theta} \) gives:

\( u' = 3^{\theta^2 + 4\theta} \left( 1 + \theta \ln(3)(2\theta + 4) \right) \)

Now substituting \( u' \) back into the derivative of \( f \):

\( f'(\theta) = \cos(\theta \cdot 3^{\theta^2 + 4\theta}) \cdot 3^{\theta^2 + 4\theta} \left( 1 + \theta \ln(3)(2\theta + 4) \right) \)

In conclusion, the derivative of the function is:

\( f'(\theta) = \cos(\theta \cdot 3^{\theta^2 + 4\theta}) \cdot 3^{\theta^2 + 4\theta} \left( 1 + \theta \ln(3)(2\theta + 4) \right) \)

It is essential to stay organized and methodical when applying these derivative rules to ensure accuracy in your calculations.

Derivatives of General Exponential Functions Example 2

In a medical lab, the quantity of a radioactive isotope, denoted as \( r \), remaining in a sample after \( t \) hours is modeled by the function:

\( r = 150 \left( \frac{1}{3} \right)^{\frac{t}{12}} \)

To analyze how this quantity changes over time, we need to find the instantaneous rate of change, represented as \( \frac{dr}{dt} \). This involves calculating the derivative of the function. The constant \( 150 \) remains outside, and we apply the derivative rules for exponential functions. The derivative of \( \left( \frac{1}{3} \right)^{\frac{t}{12}} \) is given by:

\( \frac{d}{dt} \left( \left( \frac{1}{3} \right)^{\frac{t}{12}} \right) = \left( \frac{1}{3} \right)^{\frac{t}{12}} \cdot \ln\left( \frac{1}{3} \right) \cdot \frac{1}{12} \)

Thus, the derivative can be simplified to:

\( \frac{dr}{dt} = 150 \cdot \frac{\ln\left( \frac{1}{3} \right)}{12} \cdot \left( \frac{1}{3} \right)^{\frac{t}{12}} \)

Next, we compute the instantaneous rate of change at specific time intervals: 12 hours, 24 hours, and 48 hours. It is important to convert days into hours, where 1 day equals 24 hours and 2 days equals 48 hours.

For \( t = 12 \):

\( \frac{dr}{dt} \bigg|_{t=12} = 150 \cdot \frac{\ln\left( \frac{1}{3} \right)}{12} \cdot \left( \frac{1}{3} \right)^{1} \)

Calculating this yields approximately \( -4.58 \).

For \( t = 24 \):

\( \frac{dr}{dt} \bigg|_{t=24} = 150 \cdot \frac{\ln\left( \frac{1}{3} \right)}{12} \cdot \left( \frac{1}{3} \right)^{2} \)

This results in approximately \( -1.53 \).

For \( t = 48 \):

\( \frac{dr}{dt} \bigg|_{t=48} = 150 \cdot \frac{\ln\left( \frac{1}{3} \right)}{12} \cdot \left( \frac{1}{3} \right)^{4} \)

Here, the value is approximately \( -0.17 \).

Interpreting these results reveals that the quantity of the radioactive isotope is consistently decreasing over time, as indicated by the negative values. Furthermore, the magnitude of the rate of change is diminishing, suggesting that while the isotope is still decaying, it does so at a slower rate as time progresses. This understanding is crucial in applications involving radioactive materials, as it informs safety protocols and the timing of tests.

Derivative of the Natural Exponential Function (e^x)

In calculus, understanding how to differentiate exponential functions is crucial, particularly when dealing with the natural exponential function, \( e^x \). The derivative of \( e^x \) can be derived from the general rule for exponential functions, which states that the derivative of \( b^x \) is given by:

\[\frac{d}{dx} b^x = b^x \ln(b)\]

For the specific case of \( e^x \), where the base \( b \) is \( e \), the derivative simplifies significantly because the natural logarithm of \( e \) is 1. Thus, the derivative of \( e^x \) is:

\[\frac{d}{dx} e^x = e^x\]

This property means that the function \( e^x \) is unique in that it is its own derivative. For example, if we consider a function like \( 4e^x \), we can apply the constant multiple rule, leading to:

\[\frac{d}{dx} (4e^x) = 4 \cdot \frac{d}{dx} e^x = 4e^x\]

When dealing with more complex functions involving \( e^x \), such as \( f(x) = 3e^{2x+4} \), we must apply the chain rule. The chain rule states that if a function is composed of another function, the derivative is the derivative of the outer function multiplied by the derivative of the inner function. Thus, for \( f(x) \), we have:

\[f'(x) = 3 \cdot e^{2x+4} \cdot \frac{d}{dx}(2x+4) = 3 \cdot e^{2x+4} \cdot 2 = 6e^{2x+4}\]

Another example involves the product of a polynomial and an exponential function, such as \( g(x) = x e^{5x} \). Here, we utilize the product rule, which states:

\[\frac{d}{dx}(u \cdot v) = u'v + uv'\]

Applying this to \( g(x) \), where \( u = x \) and \( v = e^{5x} \), we find:

\[g'(x) = \frac{d}{dx}(x) \cdot e^{5x} + x \cdot \frac{d}{dx}(e^{5x})\]

Calculating each part, we have:

\[g'(x) = 1 \cdot e^{5x} + x \cdot e^{5x} \cdot 5 = e^{5x} + 5xe^{5x}\]

Factoring out \( e^{5x} \) gives us:

\[g'(x) = e^{5x}(5x + 1)\]

These examples illustrate the application of the derivative rules for exponential functions, including the constant multiple rule, chain rule, and product rule, which are essential for solving more complex differentiation problems in calculus.

Find the derivative of the given function.
$f(x)=-3e^x+5x-2$

• A. $e^x+5$
• B. $-3e^x+5$
• C. $-3e^{x}\ln3+5$
• D. $-3e^x+5x-2$

Answer: (B)

Find the derivative of the given function.
$g(x)=7e^x+2x^3$

• A. $7e^x+6x^2$
• B. $7e^x+6x$
• C. $7+6x^2$
• D. $7e^x+2x^3$

Answer: (A)

Derivative of the Natural Exponential Function (e^x) Example 3

To find the derivative of the function \( y = e^{\sqrt{x^3 - 2x}} \), we will utilize the chain rule, which is essential for differentiating composite functions. The chain rule states that if you have a function \( f(g(x)) \), the derivative is given by \( f'(g(x)) \cdot g'(x) \).

Starting with the outer function, which is \( e^u \) where \( u = \sqrt{x^3 - 2x} \), we know that the derivative of \( e^u \) is simply \( e^u \). Therefore, we have:

\( \frac{dy}{dx} = e^{\sqrt{x^3 - 2x}} \cdot \frac{d}{dx}(\sqrt{x^3 - 2x}) \)

Next, we need to differentiate the inner function \( \sqrt{x^3 - 2x} \). This can be rewritten as \( (x^3 - 2x)^{1/2} \). Applying the power rule, we bring down the exponent \( \frac{1}{2} \) and decrease the exponent by one:

\( \frac{d}{dx}(\sqrt{x^3 - 2x}) = \frac{1}{2}(x^3 - 2x)^{-1/2} \cdot \frac{d}{dx}(x^3 - 2x) \)

Now, we differentiate \( x^3 - 2x \) using the power rule:

\( \frac{d}{dx}(x^3 - 2x) = 3x^2 - 2 \)

Substituting this back into our expression gives:

\( \frac{d}{dx}(\sqrt{x^3 - 2x}) = \frac{1}{2}(x^3 - 2x)^{-1/2} \cdot (3x^2 - 2) \)

Now, we can combine everything to find the full derivative:

\( \frac{dy}{dx} = e^{\sqrt{x^3 - 2x}} \cdot \frac{1}{2}(x^3 - 2x)^{-1/2} \cdot (3x^2 - 2) \)

To simplify, we can express \( (x^3 - 2x)^{-1/2} \) as \( \frac{1}{\sqrt{x^3 - 2x}} \). Thus, the derivative can be rewritten as:

\( \frac{dy}{dx} = \frac{e^{\sqrt{x^3 - 2x}}(3x^2 - 2)}{2\sqrt{x^3 - 2x}} \)

This final expression represents the derivative of the function, demonstrating the application of the chain rule and the power rule effectively. If you have any questions or need further clarification, feel free to ask!

Find the derivative of the given function.
$y=x^2e^{3x^2+5x}$

• A. $2xe^{3x^2+5x-1}$
• B. $2xe^{3x^2+5x}+x^2e^{3x^2+5x}$
• C. $2xe^{3x^2+5x-1}(6x+5)$
• D. $x{e}^{3x^2+5}(6x^2+5x+2)$

Answer: (D)

Find the derivative of the given function.
$f(t)=\frac{e^{3t}}{t-2e^{-t}}$

• A. $\frac{e^{3t}(3t-8e^{-t}-1)}{(t-2e^{-t})^2}$
• B. $\frac{3e^{3t}}{1+2e^{-t}}$
• C. $\frac{(t-1)e^{3t}}{(t-2e^{-t})^2}$
• D. $\frac{3t^2e^{3t-1}-8te^{2t-1}-e^{3t}}{(t-2e^{-t})^2}$

Answer: (A)

Find the derivative of the given function.
$y=\tan \left(e^{-x^3}\right)$

• A. $\sec^2\left(e^{-x^3}\right)$
• B. $e^{-x^3}\cdot\sec^2\left(e^{-x^3}\right)$
• C. $-3x^2\cdot e^{-x^3}\cdot\sec^2\left(e^{-x^3}\right)$
• D. $-x^3\cdot e^{-x^3-1}\cdot\sec^2\left(e^{-x^3}\right)$

Answer: (C)

Derivative of the Natural Exponential Function (e^x) Example 4

To find the equation of the tangent line to the function \( y = 3e^x - x \) at the point where \( x = 0 \), we first need to determine the slope of the tangent line, which is given by the derivative of the function. The derivative, denoted as \( \frac{dy}{dx} \), can be calculated using the rules of differentiation.

For the function \( y = 3e^x - x \), we differentiate each term separately. The derivative of \( 3e^x \) is \( 3e^x \), and the derivative of \( -x \) is \( -1 \). Therefore, the derivative of the function is:

\[\frac{dy}{dx} = 3e^x - 1\]

Next, we evaluate the derivative at \( x = 0 \) to find the slope of the tangent line at that point:

\[\frac{dy}{dx} \bigg|_{x=0} = 3e^0 - 1 = 3 \cdot 1 - 1 = 2\]

Now that we have the slope \( m = 2 \), we need to find the corresponding \( y \)-coordinate at \( x = 0 \) to use in the point-slope form of the line. We substitute \( x = 0 \) into the original function:

\[y = 3e^0 - 0 = 3 \cdot 1 - 0 = 3\]

Thus, the point of tangency is \( (0, 3) \). We can now use the point-slope form of the equation of a line, which is given by:

\[y - y_1 = m(x - x_1)\]

Substituting \( m = 2 \), \( x_1 = 0 \), and \( y_1 = 3 \) into the equation, we have:

\[y - 3 = 2(x - 0)\]

Rearranging this equation gives us:

\[y = 2x + 3\]

Therefore, the equation of the tangent line to the function \( y = 3e^x - x \) at \( x = 0 \) is:

\[y = 2x + 3\]

Derivatives of General Logarithmic Functions

To find the derivative of the logarithmic function \( y = \log_b(x) \), we can utilize the relationship between logarithmic and exponential functions. By rewriting the logarithmic equation in its exponential form, we have \( x = b^y \). This transformation allows us to apply the rules of differentiation effectively.

To differentiate both sides, we start with the left side, where the derivative of \( x \) with respect to \( x \) is simply 1. For the right side, we apply the chain rule to differentiate \( b^y \). The derivative of \( b^y \) is given by \( b^y \cdot \ln(b) \), where \( \ln(b) \) is the natural logarithm of the base \( b \). Since \( y \) is a function of \( x \), we multiply by \( \frac{dy}{dx} \) to account for the chain rule. This gives us the equation:

\( 1 = b^y \cdot \ln(b) \cdot \frac{dy}{dx} \)

Next, we isolate \( \frac{dy}{dx} \) by dividing both sides by \( b^y \cdot \ln(b) \), leading to:

\( \frac{dy}{dx} = \frac{1}{b^y \cdot \ln(b)} \)

Substituting back \( y = \log_b(x) \), we can express \( b^y \) as \( x \). Thus, the derivative simplifies to:

\( \frac{dy}{dx} = \frac{1}{x \cdot \ln(b)} \)

This formula indicates that the derivative of \( \log_b(x) \) is \( \frac{1}{x \cdot \ln(b)} \), where \( b > 0 \) and \( b \neq 1 \), and \( x > 0 \) since logarithms are only defined for positive values.

For example, to find the derivative of \( g(x) = \log_8(x) \), we apply the rule:

\( g'(x) = \frac{1}{x \cdot \ln(8)} \)

In another case, if we need to differentiate \( f(x) = \log_5(x^2) \), we recognize that this involves the chain rule. The derivative is calculated as follows:

First, we apply the logarithmic derivative rule:

\( f'(x) = \frac{1}{x^2 \cdot \ln(5)} \)

Then, we multiply by the derivative of the inside function \( x^2 \), which is \( 2x \):

\( f'(x) = \frac{1}{x^2 \cdot \ln(5)} \cdot 2x = \frac{2}{x \cdot \ln(5)} \)

Through these examples, we see how to effectively apply the derivative rules for logarithmic functions, including the necessary use of the chain rule when dealing with composite functions.

Find the derivative of the given function.
$f(x)=8\log_2x$

• A. $\frac{8}{2\ln x}$
• B. $\frac{8}{x\ln2}$
• C. $\frac{8}{x}$
• D. $\frac{1}{x\ln8}$

Answer: (B)

Find the derivative of the given function.
$g(t)=2t+\log_5t$

• A. $\frac{2}{t\ln5}$
• B. $\frac{1}{t\ln5}$
• C. $2+5\ln t$
• D. $2+\frac{1}{t\ln5}$

Answer: (D)

Derivatives of General Logarithmic Functions Example 5

To find the derivative of the function \( y = \log_4\left(\frac{\sqrt{x}}{4x + 1}\right) \), we will utilize several calculus rules, including the chain rule and the quotient rule. The derivative of a logarithmic function is given by the formula:

\[\frac{d}{dx} \log_b(u) = \frac{1}{u \ln(b)} \cdot \frac{du}{dx}\]

In this case, \( u = \frac{\sqrt{x}}{4x + 1} \) and \( b = 4 \). Thus, the derivative can be expressed as:

\[\frac{dy}{dx} = \frac{1}{\frac{\sqrt{x}}{4x + 1} \ln(4)} \cdot \frac{d}{dx}\left(\frac{\sqrt{x}}{4x + 1}\right)\]

Next, we need to differentiate \( u \) using the quotient rule, which states:

\[\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}\]

Here, let \( f(x) = \sqrt{x} \) and \( g(x) = 4x + 1 \). The derivatives are:

\[f'(x) = \frac{1}{2\sqrt{x}} \quad \text{and} \quad g'(x) = 4\]

Applying the quotient rule gives us:

\[\frac{d}{dx}\left(\frac{\sqrt{x}}{4x + 1}\right) = \frac{(4x + 1)\left(\frac{1}{2\sqrt{x}}\right) - \sqrt{x}(4)}{(4x + 1)^2}\]

Now, simplifying the numerator:

\[(4x + 1)\left(\frac{1}{2\sqrt{x}}\right) - 4\sqrt{x} = \frac{4x + 1}{2\sqrt{x}} - 4\sqrt{x} = \frac{4x + 1 - 8x}{2\sqrt{x}} = \frac{1 - 4x}{2\sqrt{x}}\]

Thus, we have:

\[\frac{d}{dx}\left(\frac{\sqrt{x}}{4x + 1}\right) = \frac{\frac{1 - 4x}{2\sqrt{x}}}{(4x + 1)^2} = \frac{1 - 4x}{2\sqrt{x}(4x + 1)^2}\]

Substituting this back into our derivative expression, we get:

\[\frac{dy}{dx} = \frac{1}{\frac{\sqrt{x}}{4x + 1} \ln(4)} \cdot \frac{1 - 4x}{2\sqrt{x}(4x + 1)^2}\]

Now, simplifying further, we can multiply through by the reciprocal of \( \frac{\sqrt{x}}{4x + 1} \):

\[\frac{dy}{dx} = \frac{(1 - 4x)(4x + 1)}{2\sqrt{x}(4x + 1)^2 \ln(4) \cdot \sqrt{x}} = \frac{(1 - 4x)}{2\ln(4)(4x + 1)\sqrt{x}}\]

Finally, the derivative of the function \( y = \log_4\left(\frac{\sqrt{x}}{4x + 1}\right) \) is:

\[\frac{dy}{dx} = \frac{1 - 4x}{2\ln(4)(4x + 1)\sqrt{x}}\]

This result encapsulates the application of logarithmic differentiation, the chain rule, and the quotient rule, demonstrating the complexity of derivatives involving composite functions.

Find the derivative of the given function.
$f(x)=(x^3+2x)\cdot\log_5x$

• A. $\frac{(3x^2+2)}{x\ln5}$
• B. $(3x^2+2)\cdot\log_5x+x^2+2$
• C. $\left(3x^2+2+\frac{x^2}{\ln5}+\frac{2}{\ln5}\right)\cdot\log_5x$
• D. $\frac{x^3+2x}{x\ln5}+\log_5x\cdot\left(3x^2+2\right)$

Answer: (D)

Find the derivative of the given function.
$g\left(t\right)=\log_5\left(7^{t^2+4}\right)$

• A. $\frac{2t\ln7}{\ln5}$
• B. $2t\ln\frac75$
• C. $\frac{2t\ln7}{t\ln5}$
• D. $\frac{7^{t^{2+4}}}{t\ln5}$

Answer: (A)

Find the derivative of the given function.
$h\left(x\right)=\sin\left(\log_4\left(x^2\right)\right)$

• A. $\frac{\cos\left(\log_4x^2\right)}{x^2\ln4}$
• B. $\frac{2\cos\left(\log_4x^2\right)}{x\ln4}$
• C. $\cos\left(\log_4x^2\right)$
• D. $\frac{2\cos\left(\log_4x^2\right)}{\ln4}$

Answer: (B)

Derivative of the Natural Logarithmic Function

In calculus, understanding the derivative of logarithmic functions is essential, particularly the natural logarithm, denoted as ln(x) or log_e(x). The derivative of the natural logarithm can be derived from the general logarithmic function's derivative rule. Specifically, the derivative of ln(x) is given by:

\[ \frac{d}{dx} \ln(x) = \frac{1}{x} \]

This result holds true under the condition that x > 0. The natural logarithm is a special case of logarithmic functions where the base is the mathematical constant e, approximately equal to 2.71828. Since ln(e) equals 1, this simplifies our derivative to 1/x.

When dealing with derivatives involving constants, such as finding the derivative of 6ln(x), the constant multiple rule applies. This means:

\[ \frac{d}{dx} [6 \ln(x)] = 6 \cdot \frac{1}{x} = \frac{6}{x} \]

For more complex functions, such as ln(x² + 4x), the chain rule must be utilized. The chain rule states that when differentiating a composite function, you differentiate the outer function and multiply it by the derivative of the inner function. Thus, the derivative of f(x) = ln(g(x)) is:

\[ f'(x) = \frac{1}{g(x)} \cdot g'(x) \]

Applying this to our function:

1. The outer function is ln, so its derivative is 1/(x² + 4x).

2. The inner function g(x) = x² + 4x has a derivative of g'(x) = 2x + 4.

Combining these results gives:

\[ f'(x) = \frac{2x + 4}{x² + 4x} \]

In another example, to find the derivative of g(x) = x \cdot ln(x³), the product rule is necessary. The product rule states that for two functions u(x) and v(x), the derivative is:

\[ (uv)' = u'v + uv' \]

Here, let u = x and v = ln(x³). The derivative of u is u' = 1, and for v, we apply the chain rule:

1. The outer function is ln, giving 1/(x³).

2. The inner function x³ has a derivative of 3x².

Thus, the derivative of v is:

\[ v' = \frac{3x²}{x³} = \frac{3}{x} \]

Applying the product rule results in:

\[ g'(x) = x \cdot \frac{3}{x} + ln(x³) \cdot 1 = 3 + ln(x³) \]

These examples illustrate the application of derivative rules, including the constant multiple rule, chain rule, and product rule, to find derivatives of logarithmic functions effectively.

Find the derivative of the given function.
$f(x)=2x^3-1+\ln x$

• A. $6x$
• B. $\frac{3x^2-1}{x}$
• C. $6x^2+\frac{1}{x}$
• D. $6x-1$

Answer: (C)

Find the derivative of the given function.
$g(x)=e^{x}+\ln x^5$

• A. $\frac{5e^{x}}{x}$
• B. $e^{x}+5\ln x$
• C. $\frac{e^{x}+5}{x}$
• D. $e^{x}+\frac{5}{x}$

Answer: (D)

Find the derivative of the given function.
$h\left(x\right)=\ln\left(\frac{\sqrt{x+1}}{x^2+3}\right)$

• A. $\frac{-3x^2-4x+3}{2\left(x+1\right)\left(x^2+3\right)}$
• B. $\frac{x^2+3}{\sqrt{x+1}}$
• C. $\frac{-3x^2-4x+3}{2x\sqrt{x+1}\cdot\left(x^2+3\right)^2}$
• D. $\frac{-3x^2-4x+3}{2\left(x+1\right)\left(x^2+3\right)}\cdot\ln\left(\frac{\sqrt{x+1}}{x^2+3}\right)$

Answer: (A)

Find the derivative of the given function.
$y=x^2\ln(x^2)$

• A. $2x\ln x^2+\frac{2}{x}$
• B. $2x\left(\ln\left(x^2\right)+1\right)$
• C. $\frac{2}{x}$
• D. $\frac{1}{x^2}$

Answer: (B)

Find the derivative of the given function.
$g(x)=e^{x^2\ln x}$

• A. $e^{x^2\ln x}$
• B. $e^{2x+\frac{1}{x}}$
• C. $xe^{x^2\ln x}\left(2\ln x+1\right)$
• D. $2e^{x^2\ln x}$

Answer: (C)

Derivative of the Natural Logarithmic Function Example 6

To find the derivative \(\frac{dy}{dx}\) using implicit differentiation, we start with the equation:

\(\ln(x^2 y) = 2e^{3x + y}\)

Taking the derivative of both sides with respect to \(x\), we apply the chain rule and product rule where necessary. The left-hand side becomes:

\(\frac{1}{x^2 y} \cdot \frac{d}{dx}(x^2 y)\)

Using the product rule on \(x^2 y\), we have:

\(x^2 \cdot \frac{dy}{dx} + y \cdot 2x\)

Thus, the left-hand side simplifies to:

\(\frac{1}{x^2 y} \left( x^2 \frac{dy}{dx} + 2xy \right)\)

For the right-hand side, we differentiate \(2e^{3x + y}\) as follows:

\(2e^{3x + y} \cdot (3 + \frac{dy}{dx})\)

Setting both sides equal gives us:

\(\frac{1}{x^2 y} \left( x^2 \frac{dy}{dx} + 2xy \right) = 2e^{3x + y} (3 + \frac{dy}{dx})\)

Next, we rearrange the equation to isolate terms involving \(\frac{dy}{dx}\). This leads to:

\(\frac{1}{y} \frac{dy}{dx} - 2e^{3x + y} \frac{dy}{dx} = 6e^{3x + y} - \frac{2}{x}\)

Factoring out \(\frac{dy}{dx}\) gives:

\(\frac{dy}{dx} \left( \frac{1}{y} - 2e^{3x + y} \right) = 6e^{3x + y} - \frac{2}{x}\)

To isolate \(\frac{dy}{dx}\), we divide both sides by \(\left( \frac{1}{y} - 2e^{3x + y} \right)\):

\(\frac{dy}{dx} = \frac{6e^{3x + y} - \frac{2}{x}}{\frac{1}{y} - 2e^{3x + y}}\)

To simplify further, we find a common denominator for the numerator and denominator. The numerator becomes:

\(6x e^{3x + y} - 2\) (after multiplying by \(x\))

And the denominator becomes:

\(1 - 2y e^{3x + y}\) (after multiplying by \(y\)).

Thus, we rewrite \(\frac{dy}{dx}\) as:

\(\frac{dy}{dx} = \frac{(6x e^{3x + y} - 2) y}{1 - 2y e^{3x + y}}\)

Finally, distributing gives us the complete expression for \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{6xy e^{3x + y} - 2y}{1 - 2y e^{3x + y}}\)

This process illustrates the importance of careful differentiation and algebraic manipulation in implicit differentiation problems.

Derivative of the Natural Logarithmic Function Example 7

In this problem, we are given a function that describes the time t in days it takes for the number of bacteria in a sample to reach a certain amount b. The function is defined as:

\( t(b) = 42 \ln\left(\frac{650}{1500 - b}\right) \)

To find the rate of change of time with respect to the number of bacteria, we need to compute the derivative of this function, denoted as \( t'(b) \). The derivative can be calculated using the chain rule and properties of logarithms. The derivative of the natural logarithm function is given by:

\( \frac{d}{dx} \ln(x) = \frac{1}{x} \)

Applying this to our function, we have:

\( t'(b) = 42 \cdot \frac{1}{1500 - b} \cdot \frac{d}{db}(1500 - b) \)

Since the derivative of \( 1500 - b \) is \(-1\), we can simplify this to:

\( t'(b) = \frac{42}{1500 - b} \)

Next, we substitute \( b = 1000 \) into the derivative to find \( t'(1000) \):

\( t'(1000) = \frac{42}{1500 - 1000} = \frac{42}{500} = 0.084 \)

This value, \( 0.084 \), represents the rate of change of time with respect to the number of bacteria when there are 1,000 bacteria present. Specifically, it indicates that at this population level, it takes approximately 0.084 days for the sample to grow by one additional bacterium. To put this into perspective, converting this rate into hours shows that it takes about 2 hours for each new bacterium to grow, which can be a more intuitive understanding of the growth rate.

In summary, the derivative \( t'(b) \) provides insight into how quickly the time required for bacterial growth changes as the population increases, highlighting the relationship between time and bacterial growth dynamics.

Graphical Applications of Exponential & Logarithmic Derivatives: Example 8

To find the critical points of the function \( y = \ln(x) - 1 \), we first need to determine its derivative. The derivative of the natural logarithm function is given by:

\( y' = \frac{1}{x} \)

In this case, since we are looking for critical points, we need to identify where the derivative is either equal to zero or does not exist. However, the expression \( \frac{1}{x} \) can never equal zero because the numerator is always 1. Therefore, we focus on where the derivative does not exist, which occurs when the denominator is zero.

Setting the denominator equal to zero gives us:

\( x = 0 \)

However, we must also consider the domain of the original function. The natural logarithm function, \( \ln(x) \), is only defined for \( x > 0 \). Therefore, \( x = 0 \) is not within the domain of the function, and we cannot consider it a critical point.

Next, we check for any other potential critical points by examining the behavior of the function. The only point where the derivative does not exist is at \( x = 1 \), but substituting \( x = 1 \) into the original function results in \( \ln(1) - 1 = 0 - 1 = -1 \), which is valid. However, since the derivative \( y' \) does not exist at this point, we need to ensure it is within the domain. Since \( \ln(0) \) is undefined, \( x = 1 \) is not a valid critical point either.

In conclusion, for the function \( y = \ln(x) - 1 \), there are no critical points, as all potential candidates either do not exist in the domain or lead to undefined values. Understanding the relationship between the function's domain and its critical points is crucial in calculus.

Graphical Applications of Exponential & Logarithmic Derivatives: Example 9

To find the absolute maximum and minimum values of the function \( f(x) = x \ln(x) \) over the closed interval \([1, 2]\), we can apply the Extreme Value Theorem. This theorem states that a continuous function on a closed interval will attain both a maximum and a minimum value.

First, we need to identify the critical points of the function by calculating its derivative. Using the product rule, which states that the derivative of a product \( u \cdot v \) is given by \( u'v + uv' \), we differentiate \( f(x) \). Here, let \( u = x \) and \( v = \ln(x) \). Thus, the derivative \( f'(x) \) is:

\[f'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1\]

Next, we find the critical points by setting the derivative equal to zero:

\[\ln(x) + 1 = 0 \implies \ln(x) = -1 \implies x = e^{-1} = \frac{1}{e} \approx 0.37\]

However, since \( \frac{1}{e} \) is not within the interval \([1, 2]\), we will only evaluate the function at the endpoints of the interval.

Now, we evaluate \( f(x) \) at the endpoints:

For \( x = 1 \):

\[f(1) = 1 \cdot \ln(1) = 1 \cdot 0 = 0\]

For \( x = 2 \):

\[f(2) = 2 \cdot \ln(2) \approx 2 \cdot 0.693 = 1.386\]

Now we compare the values obtained:

• At \( x = 1 \), \( f(1) = 0 \)
• At \( x = 2 \), \( f(2) \approx 1.386 \)

Thus, the absolute minimum value of the function on the interval \([1, 2]\) is \( 0 \) at \( x = 1 \), and the absolute maximum value is approximately \( 1.386 \) at \( x = 2 \).

In conclusion, the global minimum is \( 0 \) at \( x = 1 \) and the global maximum is approximately \( 1.386 \) at \( x = 2 \).

Identify the open intervals on which the function is increasing or decreasing.
$f(x)=xe^{-2x}$

• A. Increasing on $\left(-\infty,\frac12\right)$, Decreasing on $\left(\frac12,\infty\right)$
• B. Increasing on $\left(\frac12,\infty\right)$$\left(-\infty,\frac12\right)$, Decreasing on $\left(-\infty,\frac12\right)$
• C. Increasing on $\left(-\infty,\infty\right)$
• D. Decreasing on $\left(-\infty,\infty\right)$

Answer: (A)

Identify the local minimum and maximum values of the given function, if any.
$f(t)=t^2\ln t$, $t>0$

• A. Local maximum of $1$ at $x=1$, Local minimum of $-0.19$ at $x=\frac12$
• B. Local maximum of $-\frac{1}{2e}$ at $x=\frac{1}{\sqrt{e}}$ , No local minima
• C. Local minimum of $-\frac{1}{2e}$ at $x=\frac{1}{\sqrt{e}}$ , No local maxima
• D. No local extrema

Answer: (C)

For the following graph, find the open intervals for which the function is concave up or concave down. Identify any inflection points.

• A. Concave down: $\left(0,\infty\right)$; No inflection pt
• B. Concave down: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$
• C. Concave up: $\left(-\infty,0\right)$; Concave down: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$
• D. Concave down: $\left(-\infty,\infty\right)$; No inflection pt

Answer: (A)

Determine the intervals for which the function is concave up or concave down. State the inflection points.
$f(x)=4\ln(3x^2)$

• A. Concave down: $\left(-\infty,0\right)$; Concave up: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$
• B. Concave down: $\left(-\infty,0\right)$ & $\left(0,\infty\right)$; No inflection pt
• C. Concave up: $\left(-\infty,0\right)$; Concave down: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$
• D. Concave up: $\left(-\infty,0\right)$; Concave down: $\left(0,\infty\right)$; No inflection pt

Answer: (B)

Here’s what students ask on this topic:

What is the derivative of an exponential function b^x?

The derivative of an exponential function b^x is given by the formula:

bx⋅ln(b)

where b is a positive constant and b ≠ 1. This rule is derived using the limit definition of the derivative and properties of logarithms and exponents.

How do you find the derivative of e^x?

To find the derivative of e^x, we use the rule for exponential functions. Since the base e is a special case where ln(e) = 1, the derivative simplifies to:

ex

Thus, the derivative of e^x is simply e^x.

What is the derivative of the natural logarithm function ln(x)?

The derivative of the natural logarithm function ln(x) is given by:

1x

This rule applies for x > 0, as the natural logarithm is only defined for positive values of x.

How do you apply the chain rule to find the derivative of ln(x² + 4x)?

To find the derivative of ln(x² + 4x) using the chain rule, follow these steps:

1. Identify the outer function (ln) and the inner function (x² + 4x).

2. Differentiate the outer function: 1x2+4x

3. Multiply by the derivative of the inner function: 2x + 4.

The final derivative is:

2x+4x2+4x

What is the derivative of log_b(x)?

The derivative of the logarithmic function log_b(x) is given by:

1x⋅ln(b)

This rule applies for x > 0 and b > 0, b ≠ 1. It is derived using the properties of logarithms and the chain rule.