Find the indicated derivative. f ( x ) = 6 x 4 − 4 3 f(x)=6x^4-4^3 f ( x ) = 6 x 4 − 4 3 f ′ ( x ) = f^{\prime}\left(x\right)= f ′ ( x ) =

24 x 3 − 1 24x^3-1 24 x 3 − 1

Find the indicated derivative. y = 5 2 x 5 + 2 x 3 + 6 x − √ 3 4 y=\frac52x^5+2x^3+6x-\frac{\surd3}{4} y ′ = y^{\prime}=

25 2 x 4 + 6 x 2 + 6 \frac{25}{2}x^4+6x^2+6

5 x 4 + 3 x 2 + 6 5x^4+3x^2+6 5 x 4 + 3 x 2 + 6

25 2 x 4 + 6 x 2 + 5 \frac{25}{2}x^4+6x^2+5 2 25 x 4 + 6 x 2 + 5

25 2 x 5 + 6 x 3 + 6 x − 1 \frac{25}{2}x^5+6x^3+6x-1 2 25 x 5 + 6 x 3 + 6 x − 1

Find the indicated derivative. h ( t ) = 1 2 t 3 + 4 t 2 + 3 t h\left(t\right)=\frac12t^3+\frac{4}{t^2}+3\sqrt{t} h ′ ( t ) = h^{\prime}(t)=

3 2 t 2 − 8 t 3 + 3 2 t \frac32t^2-\frac{8}{t^3}+\frac{3}{2\sqrt{t}}

t 2 − 2 t 3 + 1 t t^2-\frac{2}{t^3}+\frac{1}{\sqrt{t}} t 2 − t 3 2 + t <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 1

3 2 t 2 − 8 t 3 + 3 2 t \frac32t^2-8t^3+\frac32\sqrt{t} 2 3 t 2 − 8 t 3 + 2 3 t <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

3 2 t 3 − 8 t 2 + 3 2 t \frac32t^3-\frac{8}{t^2}+\frac32\sqrt{t}

3. Techniques of Differentiation

Basic Rules of Differentiation

3. Techniques of Differentiation

Basic Rules of Differentiation: Videos & Practice Problems

Learn Concepts Practice Worksheet

Topic summary

To find derivatives efficiently, apply key rules: the derivative of a constant is 0, and the derivative of x is 1. Use the sum rule for functions like f(x) = x + 7, yielding f'(x) = 1. The constant multiple rule states that for g(x) = 8x, g'(x) = 8. The power rule, crucial for functions like xⁿ, gives f'(x) = n * x^n-1. This applies to negative and rational exponents, ensuring versatility in differentiation.

concept

Derivatives of Linear Functions

Video duration:

Derivatives of Linear Functions Video Summary

In calculus, derivatives represent the rate of change of a function, and there are efficient rules to find them without relying solely on limits. Understanding these rules can simplify the process significantly. For instance, the derivative of a constant function, such as $ g(x) = 7 $, is always zero. This is because a constant function has no slope; thus, the derivative, denoted as $ \frac{d}{dx} $, is $ 0 $.

Similarly, the derivative of the function $ f(x) = x $ is $ 1 $. This indicates that the slope of the tangent line to the function $ x $ is constant and equal to $ 1 $ at any point along the line.

When combining functions, the sum and difference rule becomes useful. This rule states that the derivative of the sum or difference of two functions is the sum or difference of their derivatives. For example, to find the derivative of $ x + 7 $, we apply this rule: $ \frac{d}{dx}(x + 7) = \frac{d}{dx}(x) + \frac{d}{dx}(7) = 1 + 0 = 1 $.

Next, when dealing with multiplication, the constant multiple rule applies. This rule indicates that when finding the derivative of a constant multiplied by a function, you can factor out the constant. For instance, for $ 8x $, the derivative is calculated as follows: $ \frac{d}{dx}(8x) = 8 \cdot \frac{d}{dx}(x) = 8 \cdot 1 = 8 $.

As functions become more complex, such as $ 8x + 7 $, you can apply these rules in combination. The derivative can be found by using the sum and difference rule along with the constant multiple rule, allowing for a systematic approach to finding derivatives of more complicated expressions.

Problem

Find the indicated derivative.
$f(x)=3x-4$
$f^{\prime}(x)=$

$0$

$1$

$2$

$3$

Problem

Find the indicated derivative.
$d/dt(-5t)$

$0$

$-5$

$5$

$5+t$

Problem

Find the indicated derivative.
$d/dt(-5x)$

$0$

$x$

$t$

$-5$

Problem

Find the indicated derivative.
$d/dx\left(\pi\right)$

$0$

$\pi$

$1$

$x$

concept

The Power Rule

Video duration:

The Power Rule Video Summary

Understanding how to find derivatives is crucial in calculus, especially when dealing with polynomial functions. One of the fundamental tools for this is the power rule, which simplifies the process of differentiation for functions of the form $ f(x) = x^n $, where $ n $ is a real number.

The power rule states that the derivative of $ x^n $ is given by:

$ \frac{d}{dx}(x^n) = n \cdot x^{n-1} $

This means you multiply the function by the exponent $ n $ and then decrease the exponent by 1. For example, if we take the derivative of $ x^6 $, applying the power rule yields:

$ \frac{d}{dx}(x^6) = 6 \cdot x^{6-1} = 6x^5 $

When functions involve addition or subtraction, the sum rule allows us to differentiate each term separately. For instance, if we have $ f(x) = x^4 + x^3 $, we can find the derivative as follows:

$ f'(x) = \frac{d}{dx}(x^4) + \frac{d}{dx}(x^3) $

Applying the power rule to each term gives:

$ f'(x) = 4x^3 + 3x^2 $

In cases where a constant multiplies a function, the constant multiple rule is applied. For example, for the function $ g(x) = 3x^2 $, the derivative can be calculated as:

$ g'(x) = 3 \cdot \frac{d}{dx}(x^2) $

Using the power rule on $ x^2 $ results in:

$ g'(x) = 3 \cdot 2x^{2-1} = 6x $

By mastering the power rule, sum rule, and constant multiple rule, you will be well-equipped to tackle a wide range of polynomial functions and their derivatives. Continued practice will enhance your proficiency in differentiation, paving the way for more complex calculus concepts.

concept

The Power Rule: Negative & Rational Exponents

Video duration:

The Power Rule: Negative & Rational Exponents Video Summary

The power rule is a fundamental tool in calculus for finding derivatives of power functions, including those with negative and rational exponents. The power rule states that if you have a function of the form $ f(x) = x^n $, the derivative $ f'(x) $ can be calculated using the formula:

\[ f'(x) = n \cdot x^{n-1} \]

Let's explore how to apply this rule to various types of exponents.

For example, consider the function $ f(x) = x^{-3} $. To find the derivative, we apply the power rule:

\[ f'(x) = -3 \cdot x^{-3-1} = -3 \cdot x^{-4} \]

This can also be expressed without a negative exponent as:

\[ f'(x) = -\frac{3}{x^4} \]

Next, let's look at a function with a rational exponent, such as $ f(x) = x^{\frac{3}{2}} $. Again, we apply the power rule:

\[ f'(x) = \frac{3}{2} \cdot x^{\frac{3}{2}-1} = \frac{3}{2} \cdot x^{\frac{1}{2}} \]

Now, consider the function $ f(x) = \frac{1}{x} $. This can be rewritten as $ f(x) = x^{-1} $ to apply the power rule:

\[ f'(x) = -1 \cdot x^{-1-1} = -1 \cdot x^{-2} \]

Rewriting this gives us:

\[ f'(x) = -\frac{1}{x^2} \]

Lastly, for the function $ f(x) = \sqrt[3]{x} $, we can express it as $ f(x) = x^{\frac{1}{3}} $ and apply the power rule:

\[ f'(x) = \frac{1}{3} \cdot x^{\frac{1}{3}-1} = \frac{1}{3} \cdot x^{-\frac{2}{3}} \]

This can also be written as:

\[ f'(x) = \frac{1}{3} \cdot \frac{1}{x^{\frac{2}{3}}} = \frac{1}{3\sqrt[3]{x^2}} \]

In summary, the power rule is versatile and can be applied to functions with negative and rational exponents by rewriting them as power functions when necessary. Mastering this rule will greatly enhance your ability to differentiate a wide range of functions.

example

The Power Rule: Negative & Rational Exponents Example 1

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The Power Rule: Negative & Rational Exponents Example 1 Video Summary

To find the derivative of the function $ y = 2x^3 - x^2 + 4x + 7 $, we will apply the rules of differentiation step by step. The derivative is denoted as $ \frac{dy}{dx} $ or $ y' $, indicating that we are differentiating with respect to $ x $.

We can break down the function into individual terms and apply the sum and difference rules of derivatives. The function consists of four terms: $ 2x^3 $, $ -x^2 $, $ 4x $, and $ 7 $. We will differentiate each term separately:

1. For the first term $ 2x^3 $, we apply the power rule, which states that the derivative of $ x^n $ is $ nx^{n-1} $. Thus, the derivative is:

\[\frac{d}{dx}(2x^3) = 2 \cdot 3x^{3-1} = 6x^2\]

2. For the second term $ -x^2 $, we again use the power rule:

\[\frac{d}{dx}(-x^2) = -2x^{2-1} = -2x\]

3. The third term is $ 4x $. The derivative of $ x $ is $ 1 $, so:

\[\frac{d}{dx}(4x) = 4 \cdot 1 = 4\]

4. Finally, the derivative of the constant $ 7 $ is $ 0 $:

\[\frac{d}{dx}(7) = 0\]

Now, we can combine all these derivatives together:

\[\frac{dy}{dx} = 6x^2 - 2x + 4 + 0\]

After simplifying, we find that the final derivative of the function is:

\[\frac{dy}{dx} = 6x^2 - 2x + 4\]

As you practice more with derivatives, you will become more proficient and may skip detailed steps, directly moving from the original function to its derivative. Continuous practice will enhance your understanding and speed in differentiation.

Problem

Find the indicated derivative.
$f(x)=6x^4-4^3$
$f^{\prime}\left(x\right)=$

$4x^3$

$24x^3$

$4x^3-1$

$24x^3-1$

Problem

Find the indicated derivative.
$d / d t (5 - 7 t^{- 3})$

$1 - 21 t^{- 4}$

$- 21 t^{- 2}$

$- 3 t^{- 4}$

$21 t^{- 4}$

Problem

Find the indicated derivative.
$y = \frac{5}{2} x^{5} + 2 x^{3} + 6 x - \frac{\sqrt 3}{4}$
$y^{'} =$

$5x^4+3x^2+6$

$\frac{25}{2}x^4+6x^2+5$

$\frac{25}{2}x^5+6x^3+6x-1$

$\frac{25}{2} x^{4} + 6 x^{2} + 6$

Problem

Find the indicated derivative.
$h (t) = \frac{1}{2} t^{3} + \frac{4}{t^{2}} + 3 \sqrt{t}$
$h^{'} (t) =$

$t^2-\frac{2}{t^3}+\frac{1}{\sqrt{t}}$

$\frac32t^2-8t^3+\frac32\sqrt{t}$

$\frac{3}{2} t^{2} - \frac{8}{t^{3}} + \frac{3}{2 \sqrt{t}}$

$\frac{3}{2} t^{3} - \frac{8}{t^{2}} + \frac{3}{2} \sqrt{t}$

example

The Power Rule: Negative & Rational Exponents Example 2

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The Power Rule: Negative & Rational Exponents Example 2 Video Summary

In this problem, we explore the function $ f(x) = 2x^3 - 3x + 5 $ and address several key concepts in calculus, including derivatives, tangent lines, and horizontal tangents.

To begin, we find the derivative of the function, denoted as $ f'(x) $. Utilizing the power rule, we differentiate each term of the polynomial. The derivative of $ 2x^3 $ is calculated by bringing down the exponent (3) and multiplying it by the coefficient (2), resulting in $ 6x^2 $. The next term, $ -3x $, simplifies to $ -3 $ since the derivative of $ x $ is 1. The constant term $ +5 $ has a derivative of 0. Therefore, the derivative of the function is:

$ f'(x) = 6x^2 - 3 $

Next, we determine the equation of the tangent line at the point where $ x = 1 $. To do this, we need the slope of the tangent line and a point on the line. The slope $ m $ can be found by evaluating the derivative at $ x = 1 $:

$ f'(1) = 6(1)^2 - 3 = 6 - 3 = 3 $

Now, we need to find the corresponding $ y $-value, $ y_1 $, by substituting $ x = 1 $ back into the original function:

$ f(1) = 2(1)^3 - 3(1) + 5 = 2 - 3 + 5 = 4 $

With $ m = 3 $ and the point $ (1, 4) $, we can use the point-slope form of the equation of a line, which is given by:

$ y - y_1 = m(x - x_1) $

Substituting in our values, we have:

$ y - 4 = 3(x - 1) $

Distributing and rearranging gives us the equation of the tangent line:

$ y = 3x + 1 $

Finally, we explore the conditions under which the tangent line is horizontal. A horizontal tangent line occurs when the slope is zero. Therefore, we set the derivative equal to zero:

$ 6x^2 - 3 = 0 $

Solving for $ x $, we first add 3 to both sides:

$ 6x^2 = 3 $

Next, we divide by 6:

$ x^2 = \frac{1}{2} $

Taking the square root of both sides yields:

$ x = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2} $

Thus, the values of $ x $ for which the tangent line is horizontal are $ x = \pm \frac{\sqrt{2}}{2} $.

In summary, we have successfully derived the function's derivative, determined the equation of the tangent line at a specific point, and identified the conditions for horizontal tangents, reinforcing our understanding of these fundamental calculus concepts.

example

The Power Rule: Negative & Rational Exponents Example 3

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The Power Rule: Negative & Rational Exponents Example 3 Video Summary

In this example, we explore the application of derivatives in a real-world context, specifically focusing on cell culture growth. The function representing the number of cells in a culture over time is given by p(t) = 1500 + 2t², where p is the number of cells and t is the time in hours.

To find the number of cells after 24 hours, we evaluate p(24). By substituting 24 into the function, we calculate:

p(24) = 1500 + 2(24)² = 1500 + 2(576) = 1500 + 1152 = 2652

This result indicates that after 24 hours, there are 2,652 cells present in the culture.

Next, we need to determine the derivative, p'(t), to understand the rate of change of the cell population. Using the power rule for differentiation, we find:

p'(t) = d/dt(1500 + 2t²) = 0 + 4t = 4t

Now, we evaluate the derivative at t = 24:

p'(24) = 4(24) = 96

This value signifies that the rate of change of the number of cells after 24 hours is 96 cells per hour. In other words, at this point in time, the cell culture is increasing by 96 cells every hour.

Understanding these calculations helps in grasping the significance of derivatives in biological contexts, particularly how they can describe growth rates and changes over time. This critical thinking about derivatives allows for better application in various scenarios, reinforcing the importance of mathematical concepts in real-world situations.

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Basic Rules of Differentiation

3. Techniques of Differentiation

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3. Techniques of Differentiation

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Here’s what students ask on this topic:

What is the power rule for differentiation and how is it applied?

The power rule is a fundamental tool in calculus for finding the derivative of power functions. It states that if you have a function of the form $x^{n}$ , where $n$ is any real number, the derivative is $n x^{n - 1}$ . To apply the power rule, multiply the original exponent by the coefficient of $x^{n}$ and then subtract one from the exponent. For example, the derivative of $x^{6}$ is $6 x^{5}$ . This rule simplifies the process of differentiation, especially for polynomials and functions with negative or fractional exponents.

How do you find the derivative of a constant function?

The derivative of a constant function is always zero. This is because a constant function, such as $g (x) = 7$ , has a graph that is a horizontal line. The slope of a horizontal line is zero, which is why the derivative is zero. Mathematically, if $c$ is a constant, then $\frac{d d x}{c} = 0$ . This rule is useful for simplifying the differentiation of more complex functions that include constant terms.

Can the power rule be used for negative and fractional exponents?

Yes, the power rule can be applied to functions with negative and fractional exponents. The rule remains the same: for a function $x^{n}$ , the derivative is $n x^{n - 1}$ . For example, the derivative of $x^{- 3}$ is $- 3 x^{- 4}$ . Similarly, for fractional exponents, such as $x^{\frac{3}{2}}$ , the derivative is $\frac{3}{2} x^{\frac{1}{2}}$ . This flexibility makes the power rule a powerful tool in calculus.

What is the constant multiple rule in differentiation?

The constant multiple rule in differentiation states that when you have a constant multiplied by a function, you can take the constant out of the differentiation process. Mathematically, if $c$ is a constant and $f (x)$ is a function, then $\frac{d d x}{c f (x)} = c \frac{d d x}{f (x)}$ . For example, the derivative of $3 x^{2}$ is $3 (2 x) = 6 x$ . This rule simplifies the differentiation process by allowing you to focus on the function itself.

How do you differentiate a function like 1/x using the power rule?

To differentiate a function like $\frac{1}{x}$ using the power rule, you first need to rewrite it as a power function. The function $\frac{1}{x}$ can be expressed as $x^{- 1}$ . Now, apply the power rule: the derivative of $x^{- 1}$ is $- 1 x^{- 2}$ , which can be rewritten as $- \frac{1}{x^{2}}$ . This approach allows you to use the power rule for functions that initially don't appear to be power functions.