# Calculus: Derivatives III

## Tangent Lines

The derivative $f'(x)$ of a function $f(x)$ gives the slope of the tangent line $f'(x_0)$ at a point $(x_0,y_0)$ on the original function. With the slope and point, one can solve for the equation of the tangent line, $y=mx+b$. Let $y=y_0$, $m=f'(x_0)$, and $x=x_0$, and then solve for $b$.

## Problems

1. Find the equation of the line tangent to $y=4x^2 + 2$ when $x=2$. Draw a graph of the function and the tangent line, highlighting the point of intersection.

Step 1: Take the derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx}\left(4x^2 + 2\right) \\ \dfrac{dy}{dx} = 8x \\$

Step 2:Find the value of $y$ when $x=2$:

$y = 4(2)^2 + 2 \\ y = 4\cdot4 + 2 \\ y = 18 \\$

Step 3:Find the value of $\dfrac{dy}{dx}$ when $x=2$:

$\dfrac{dy}{dx} = 8(2) \\ \dfrac{dy}{dx} = 16 \\$

Step 4:Solve for the $y$-intercept of the line:

$y = mx + b \\ 18 = 16(2) + b \\ 18 = 32 + b \\ b = -14 \\$

The equation of the tangent line is $y = 16x - 14$.

Graph:

2. Find the equation of the line tangent to $y=-2x^2+1$ when $x=0$. Draw a graph of the function and the tangent line, highlighting the point of intersection.

Step 1: Take the derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx}\left(-2x^2+1\right) \\ \dfrac{dy}{dx} = -4x \\$

Step 2: Find the value of $y$ when $x=2$:

$y = -2(0)^2 + 1 \\ y = -2\cdot0 + 1 \\ y = 1 \\$

Step 3: Find the value of $\dfrac{dy}{dx}$ when $x=0$:

$\dfrac{dy}{dx} = -4\cdot 0 \\ \dfrac{dy}{dx} = 0 \\$

Step 4: Solve for the $y$-intercept of the line:

$y = mx + b \\ 1 = 0(0) + b \\ b = 1 \\$

The equation of the tangent line is $y = 1$.

Graph:

3. Find the equation of the line tangent to $y=\sin(x)$ when $x=0$. Draw a graph of the function and the tangent line, highlighting the point of intersection.

Step 1: Take the derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx}\sin(x) \\ \dfrac{dy}{dx} = \cos(x) \\$

Step 2: Find the value of $y$ when $x=0$:

$y = \sin(0) \\ y = 0 \\$

Step 3: Find the value of $\dfrac{dy}{dx}$ when $x=0$:

$\dfrac{dy}{dx} = \cos(0) \\ \dfrac{dy}{dx} = 1 \\$

Step 4: Solve for the $y$-intercept of the line:

$y = mx + b \\ 0 = 1(0) + b \\ b = 1 \\$

The equation of the tangent line is $y = x$.

Graph:

4. Find the equation of the line tangent to $y=2e^{-x}$ when $x=1$. Draw a graph of the function and the tangent line, highlighting the point of intersection.

Step 1: Take the derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx}2x^{-x} \\ \dfrac{dy}{dx} = -2e^{-x} \\$

Step 2: Find the value of $y$ when $x=1$:

$y = 2e^{-1} \\$

Step 3: Find the value of $\dfrac{dy}{dx}$ when $x=1$:

$\dfrac{dy}{dx} = -2e^{-1} \\$

Step 4: Solve for the $y$-intercept of the line:

$y = mx + b \\ 2e^{-1} = -2e^{-1}(1) + b \\ 2e^{-1} = -2e^{-1} + b \\ b = 4e^{-1} \\$

The equation of the tangent line is $y = -2e^{-1}x + 4e^{-1}$.

Graph:

5. Find the equation of the line tangent to $y=\ln(3x)-1$ when $x=1$. Draw a graph of the function and the tangent line, highlighting the point of intersection.

Step 1: Take the derivative:

$\dfrac{d}{dx}y = \dfrac{d}{dx}\ln(3x)-1 \\ \dfrac{dy}{dx} = \dfrac{1}{3x} \cdot 3 \\ \dfrac{dy}{dx} = \dfrac{1}{x} \\$

Step 2: Find the value of $y$ when $x=1$:

$y = \ln(3\cdot1)-1 \\ y = \ln(3)-1 \\$

Step 3: Find the value of $\dfrac{dy}{dx}$ when $x=1$:

$\dfrac{dy}{dx} = \dfrac{1}{1} \\ \dfrac{dy}{dx} = 1 \\$

Step 4: Solve for the $y$-intercept of the line:

$y = mx + b \\ \ln(3)-1 = 1\cdot1 + b \\ \ln(3)-1 = 1 + b \\ b = \ln(3)-2 \\$

The equation of the tangent line is $y = x + \ln(3)-2$.

Graph: