Tangent Lines & Derivatives: Videos & Practice Problems

Understanding the difference between secant and tangent lines is crucial in calculus. A secant line intersects a curve at two points, while a tangent line touches the curve at just one point, representing the instantaneous rate of change, or derivative. The derivative can be calculated using the limit as \( h \) approaches 0: $lim\left(f\right(x+h)-f(x\left)\right)/h$. This allows for finding the slope of the tangent line at any point on a function.

Slopes of Tangent Lines

In calculus, understanding the concepts of secant and tangent lines is crucial for analyzing the behavior of functions. A secant line intersects a curve at two distinct points, while a tangent line touches the curve at just one point, representing the slope at that specific location. This distinction is fundamental as it leads to different interpretations of rates of change.

To illustrate this, consider the function \( f(x) = x^2 \). To find the slope of the tangent line at \( x = 1 \), we utilize the limit definition of the derivative. The slope of the tangent line can be expressed as:

\[m = \lim_{{x \to c}} \frac{f(x) - f(c)}{x - c}\]

In this case, \( c = 1 \). Thus, we need to evaluate:

\[m = \lim_{{x \to 1}} \frac{f(x) - f(1)}{x - 1}\]

Substituting \( f(x) = x^2 \) and calculating \( f(1) \), we find:

\[f(1) = 1^2 = 1\]

Now, substituting back into the limit gives:

\[m = \lim_{{x \to 1}} \frac{x^2 - 1}{x - 1}\]

To simplify, we recognize that \( x^2 - 1 \) can be factored as \( (x - 1)(x + 1) \). Thus, we rewrite the limit as:

\[m = \lim_{{x \to 1}} \frac{(x - 1)(x + 1)}{x - 1}\]

By canceling \( (x - 1) \), we have:

\[m = \lim_{{x \to 1}} (x + 1)\]

Now, substituting \( x = 1 \) yields:

\[m = 1 + 1 = 2\]

This result indicates that the slope of the tangent line at \( x = 1 \) is 2. In contrast, if we were to calculate the slope of the secant line between two points, we would find a different value, illustrating the concept of average rate of change versus instantaneous rate of change.

In summary, the slope of the tangent line represents the instantaneous rate of change of the function at a specific point, and this concept is encapsulated in the derivative. Understanding these differences between secant and tangent lines is essential for further studies in calculus and its applications.

Given the function $f(x)=4x^2-1$, calculate the slope of the tangent line at $x=-3$.

• A. 36
• B. -36
• C. -24
• D. 24

Answer: (C)

Given the function $f\left(x\right)=x^2-10x+2$, calculate the slope of the tangent line at $x=2$.

• A. -6
• B. 6
• C. 10
• D. -10

Answer: (A)

Given the function $f\left(x\right)=x^2+100$, calculate the slope of the tangent line at $x=0$.

• A. 0
• B. 100
• C. -100
• D. Infinite (vertical line)

Answer: (A)

Equations of Tangent Lines

Understanding tangent lines is crucial in calculus, as they represent the slope of a function at a specific point. A tangent line touches the curve of a function at exactly one point, and calculating its slope involves using limits. To find the equation of a tangent line, we can follow a systematic approach that builds on previously learned concepts.

Consider the function \( f(x) = 3x^2 - 4 \) and the task of finding the equation of the tangent line at \( x = -2 \). The first step is to determine the corresponding \( y \)-value, \( f(c) \), where \( c = -2 \). By substituting \( -2 \) into the function, we calculate:

\( f(-2) = 3(-2)^2 - 4 = 3(4) - 4 = 12 - 4 = 8 \).

Now that we have \( c = -2 \) and \( f(c) = 8 \), we can proceed to find the slope of the tangent line using the limit definition of the derivative:

\( m = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \lim_{x \to -2} \frac{f(x) - 8}{x + 2} \).

Substituting \( f(x) = 3x^2 - 4 \) into the equation gives:

\( m = \lim_{x \to -2} \frac{3x^2 - 4 - 8}{x + 2} = \lim_{x \to -2} \frac{3x^2 - 12}{x + 2} \).

Next, we simplify the expression. Factoring out a 3 from the numerator results in:

\( m = \lim_{x \to -2} \frac{3(x^2 - 4)}{x + 2} = \lim_{x \to -2} \frac{3(x + 2)(x - 2)}{x + 2} \).

We can cancel \( x + 2 \) from the numerator and denominator, leading to:

\( m = \lim_{x \to -2} 3(x - 2) \).

Now substituting \( x = -2 \) gives:

\( m = 3(-2 - 2) = 3(-4) = -12 \).

With the slope \( m = -12 \) and the point \( (-2, 8) \), we can use the point-slope form of the equation of a line:

\( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (-2, 8) \).

Substituting the values, we have:

\( y - 8 = -12(x + 2) \).

Distributing the slope results in:

\( y - 8 = -12x - 24 \).

Finally, adding 8 to both sides yields:

\( y = -12x - 16 \).

This equation represents the tangent line to the function at the specified point. Mastering this process of finding tangent lines is essential for solving more complex calculus problems, and practice will enhance your understanding and skills in this area.

Given the function $f\left(x\right)=3(x^2-1)$, find the equation of the tangent line at $x=1$.

• A. $y=-6x+6$
• B. $y=-6x-6$
• C. $y=6x+6$
• D. $y=6x-6$

Answer: (D)

Determine Continuity Algebraically Example 1

To find the equation of the tangent line to the function \( f(x) = -4x^2 \) at the point where \( x = -2 \), we start by determining the slope of the tangent line using the limit definition. The slope \( m \) of the tangent line can be expressed as:

\[m = \lim_{{x \to c}} \frac{{f(x) - f(c)}}{{x - c}}\]

In this case, \( c = -2 \). First, we need to calculate \( f(-2) \):

\[f(-2) = -4(-2)^2 = -4(4) = -16\]

Now substituting into the slope formula, we have:

\[m = \lim_{{x \to -2}} \frac{{-4x^2 - (-16)}}{{x + 2}} = \lim_{{x \to -2}} \frac{{-4x^2 + 16}}{{x + 2}}\]

Next, we simplify the expression. Factoring the numerator gives:

\[-4(x^2 - 4) = -4(x + 2)(x - 2)\]

Thus, the limit becomes:

\[m = \lim_{{x \to -2}} \frac{{-4(x + 2)(x - 2)}}{{x + 2}}\]

We can cancel \( (x + 2) \) from the numerator and denominator (noting that \( x \neq -2 \) during the limit process), leading to:

\[m = \lim_{{x \to -2}} -4(x - 2) = -4(-2 - 2) = -4(-4) = 16\]

Now that we have the slope \( m = 16 \), we can use the point-slope form of the equation of a line, which is:

\[y - y_0 = m(x - x_0)\]

Here, \( (x_0, y_0) = (-2, -16) \). Substituting the values, we get:

\[y - (-16) = 16(x - (-2))\]

This simplifies to:

\[y + 16 = 16(x + 2)\]

Distributing the 16 gives:

\[y + 16 = 16x + 32\]

Finally, isolating \( y \) results in:

\[y = 16x + 32 - 16\]

Thus, the equation of the tangent line is:

\[y = 16x + 16\]

In summary, the slope of the tangent line at \( x = -2 \) is 16, and the equation of the tangent line is \( y = 16x + 16 \).

Derivatives

Understanding derivatives is crucial in calculus, as they represent the slope of a tangent line to a function at any given point. The derivative can be calculated using the limit definition, which states that the derivative of a function \( f(x) \) is given by:

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

In this context, \( h \) represents a small change in \( x \), allowing us to find the slope as two points on the function converge to a single point. To illustrate this, consider the function \( f(x) = x^2 \). To find its derivative, we substitute into the limit definition:

\( f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \)

Expanding \( (x+h)^2 \) gives us:

\( f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h} \)

Factoring out \( h \) from the numerator allows us to simplify:

\( f'(x) = \lim_{h \to 0} (2x + h) \)

As \( h \) approaches 0, the derivative simplifies to:

\( f'(x) = 2x \)

This result indicates that the slope of the tangent line to the curve \( f(x) = x^2 \) at any point \( x \) is \( 2x \). To find the slope at specific points, such as \( x = 1 \) and \( x = -2 \), we simply substitute these values into the derivative:

\( f'(1) = 2(1) = 2 \)

\( f'(-2) = 2(-2) = -4 \)

Thus, the slope of the tangent line at \( x = 1 \) is 2, and at \( x = -2 \) it is -4. This method allows for the calculation of the slope at any \( x \) value, demonstrating the power of derivatives in analyzing the behavior of functions.

Find the derivative of the function $f\left(x\right)=4x^2-9x$.

• A. $f^{\prime }\left(x\right)=4x-18$
• B. $f^{\prime}\left(x\right)=8x-9$
• C. $f^{\prime }\left(x\right)=0$
• D. $f^{\prime }\left(x\right)=4x-9$

Answer: (B)

Use the definition of a derivative, to find the derivative of the function $g(x)=x^3$ at $x=-1$.

• A. 0
• B. -1
• C. -3
• D. 3

Answer: (D)

Here’s what students ask on this topic:

What is the difference between a secant line and a tangent line?

A secant line intersects a curve at two points, while a tangent line touches the curve at just one point. The secant line represents the average rate of change between two points on the curve, whereas the tangent line represents the instantaneous rate of change at a single point. The slope of the tangent line at a point is known as the derivative of the function at that point.

How do you find the slope of a tangent line at a specific point?

To find the slope of a tangent line at a specific point, you use the limit definition of the derivative. The formula is: $\lim \left(\frac{f(x+h)-f\left(x\right)}{h}\right),h\to 0$. This formula calculates the slope of the tangent line by finding the limit as h approaches 0 of the difference quotient.

What is the equation of a tangent line to a function at a given point?

The equation of a tangent line to a function at a given point can be found using the point-slope form of a line: $y-y$₁=m(x-x₁), where $m$ is the slope of the tangent line (the derivative at the point), and $x$₁ and $y$₁ are the coordinates of the point of tangency.

How do you find the derivative of a function at any point?

To find the derivative of a function at any point, you use the general formula for the derivative: $f\text{'}\left(x\right)=\lim \left(\frac{f(x+h)-f\left(x\right)}{h}\right),h\to 0$. This formula calculates the derivative by finding the limit as h approaches 0 of the difference quotient, which gives the slope of the tangent line at any point x.

What is the instantaneous rate of change and how is it related to the derivative?

The instantaneous rate of change of a function at a specific point is the rate at which the function's value is changing at that exact point. It is represented by the slope of the tangent line at that point. The derivative of a function at a point gives the instantaneous rate of change of the function at that point. Therefore, the derivative is a measure of how the function's value changes instantaneously with respect to changes in the input value.