1.10 Definition of Derivative

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In this section we learn the definition of the derivative and we use it to solve the tangent line problem.

Tangent Lines

The idea of a tangent line is that if we zoom in on a point $(a, f(a))$ on the graph of a smooth function, $y = f(x)$ , then the graph looks like a straight line. This line is the tangent line and the point $(a, f(a))$ is called the point of tangency.

The interactive graph below shows a function and its tangent line at a point. You can drag the point along the curve and see how the tangent line changes. You can also zoom in on the curve to see that the tangent line approximates the curve locally. If you zoom in far enough, the curve and the tangent line become nearly indistinguishable.

The tangent line problem is to find the equation of the tangent line to the graph of $y = f(x)$ at the point $(a, f(a))$ . From the point-slope form of the equation of a line, the equation of the tangent line has the form $y -f(a) = m(x-a),$ where $m$ is the slope of the line. The derivative, denoted $f'(x)$ and read ‘f prime of x’, is the mathematical function which will give us the slope of the tangent line at any point. In particular, the slope of the tangent line at the point $(a, f(a))$ is denoted by $f'(a)$ . Thus, using $m = f'(a)$ , the equation of the tangent line can be written as $y - f(a) = f'(a)(x-a), \quad \text {or} \quad y = f(a) + f'(a)(x-a).$

We will find the second form useful later when we study linear approximation. By distributing and adding, we can convert the point-slope form into the more common slope-intercept form, $y = mx + b$ . For example, if the point is $(1,2)$ and $m = 3$ , then the point-slope form of the line is $y - 2 = 3(x - 1).$ Distributing the 3 and then adding 2, gives the slope intercept form: $y- 2 = 3x-3$ $y = 3x - 1.$

The Derivative

Our objective is to discover a method for computing the derivative, $f'(x)$ , of a given function, $y = f(x)$ . Since the derivative represents the slope of the tangent line to the graph of a function at a point, we begin by recalling the formula for the slope of a line between two points, $(x_1, y_1)$ and $(x_2, y_2)$ : $m = \frac {\text {rise}}{\text {run}} = \frac {\Delta y}{\Delta x} = \frac {y_2 - y_1}{x_2 - x_1}.$

To find the slope of the tangent line to the graph of $y = f(x)$ at the general point $(x, f(x))$ , we first compute the slope of the secant line between the point $(x, f(x))$ and a “nearby” point $(x+h, f(x+h))$ : $m_{\text {sec}} = \frac {f(x+h) - f(x)}{(x+h)-x}.$ We can simplify the denominator: $m_{\text {sec}} = \frac {f(x+h) - f(x)}{h}.$ This quantity is known as the difference quotient.

The slope of the tangent line, is obtained by letting $h \to 0$ which has the effect of moving the point $(x+h, f(x+h))$ towards the point $(x, f(x))$ .

This gives us the definition of the derivative:

Derivative The derivative of the function $f(x)$ , denoted by $f'(x)$ is defined by $f'(x) = \lim _{h \to 0} \frac {f(x+h)-f(x)}{h}.$ This definition is valid for all values of $x$ for which the limit exists.

Other notations for the derivative include $y',\frac {dy}{dx}$ and $\frac {d}{dx} f(x)$ .

Below is an interactive graph that shows both a tangent line (in red) and a secant line (in green). Consider the red dot to be the point $(x, f(x))$ . Move the red dot to a point of your choosing. The green dot represents the point $(x+h, f(x+h))$ . Move the green dot toward the red dot. Observe that when the green dot is very close to the red dot (i.e. $h$ is very small), the secant line becomes indistinguishable from the tangent line. If the green dot is to the right of the red dot, then $h$ is positive, and if the green dot is to the left of the red dot, then $h$ is negative.

We now compute the derivative of several different functions.

Computing the derivative from the definition

example 1 Compute the derivative of $f(x) = x^2$ and find $f'(-3), f'(0)$ and $f'(2)$ . Using the definition of the derivative, we have $\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{(x+h)^2- x^2}{h}\\[5pt] &= \lim_{h \to 0} \frac{(x^2 + 2xh + h^2)- x^2}{h} \\[5pt] &= \lim_{h \to 0} \frac{ (2xh + h^2)}{h}\\[5pt] &= \lim_{h \to 0} \frac{ h(2x + h)}{h}\\[5pt] &= \lim_{h \to 0} (2x + h)\\ &= 2x. \end{align*}$

Thus the derivative of $x^2$ is $2x$ and this tells use the slope of the tangent line at a given $x$ -value. The process of obtaining the derivative is called differentiation, and the mathematical symbol for the differentiation operator is $\frac {d}{dx}$ so that we can write: $\frac {d}{dx}\left ( x^2\right ) = 2x.$ This is essentially the notation used by German polymath, philosopher and co-discoverer of calculus, Gottfried Wilhelm Leibniz circa 1700. Now let’s use the derivative to discuss tangent lines and their slopes.

At the point $(-3, 9)$ on the graph of the parabola, $y = x^2$ , the slope of the tangent line is given by $f'(-3) = 2x\big |_{x=-3} = 2(-3) = -6.$ That is, we plug the x-coordinate, $x=-3$ , into the function $f'(x) = 2x$ .
The points $(0,0)$ and $(2,4)$ are also on the parabola. At the point $(0,0)$ , the slope of the tangent line is $m_{\text {tan}}= f'(0) =2x\big |_{x=0}= 2(0) = 0,$ and at the point $(2, 4)$ the slope is $m_{\text {tan}}= f'(2) =2x\big |_{x=2}= 2(2) = 4.$

The graph below shows the parabola, $y = x^2$ , along with the tangent lines at $x = -3, 0,$ and $2$ .

(problem 1) Use the fact that the derivative of $x^2$ is $2x$ , i.e., $\frac {d}{dx}\left ( x^2 \right ) = 2x,$ to find the equation of the tangent line to the graph of $f(x) = x^2$ at the point where $x = 1/2$ .

The point of tangency is $(1/2, \answer {1/4})$ .

The point of tangency is a point on the graph of $y = f(x)$

The slope of the tangent line is $\answer {1}$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(1/2) = 2x\big |_{x=1/2}$

The equation of the tangent line is $y = \; \answer {x - 1/4}$ .

The point-slope form of the equation of a line is given by $y - y_1 = m(x - x_1).$

example 2 Compute the derivative, $f'(x)$ , for the function $f(x) = x^2 - 5x + 7$ .
Step 1. Compute $f(x+h)$ .
Substitute $(x+h)$ into the formula for $f(x)$ in the place of $x$ : $\begin{align*} f(x+h) &= (x+h)^2 - 5(x+h) + 7 \\ &= x^2 + 2xh + h^2 - 5x -5h + 7. \end{align*}$

Step 2. Compute the numerator of the difference quotient, $f(x+h) - f(x)$ .
Using the result of step 1, we have $f(x+h) - f(x) = \left (x^2 + 2xh + h^2 - 5x -5h + 7\right ) - \left (x^2 - 5x + 7\right )$ $= x^2 + 2xh + h^2 - 5x - 5h + 7 -x^2 + 5x - 7 = 2xh + h^2 - 5h.$

Step 3. Divide by $h$ to obtain the difference quotient.
Using the result of step 2, $\frac {f(x+h) - f(x)}{h} = \frac {2xh + h^2 - 5h}{h} = \frac {h(2x + h - 5)}{h} = 2x + h -5.$

Step 4. Compute the limit of the difference quotient.
Using step 3, we have $\lim _{h \to 0} \frac {f(x+h) - f(x)}{h} = \lim _{h \to 0} (2x + h -5) = 2x - 5.$

Thus, the derivative of $f(x) = x^2 - 5x + 7$ is $f'(x) = 2x -5$ . This can also be written using Leibniz notation as

$\frac {d}{dx} \left (x^2 - 5x + 7\right ) = 2x-5.$

In general, if y = f(x) then the notations $y' , \; \frac {dy}{dx},\; \frac {d}{dx} f(x) \; \text {and} \; f'(x),$ are all valid ways to express the derivative. Furthermore, the following are valid ways to express the derivative evaluated at a point $(a, f(a))$ , as when finding the slope of a tangent line: $y'(a) = y' \big |_{x=a} = \frac {dy}{dx}\Big |_{x=a} = \frac {d}{dx} f(x)\Big |_{x=a} = f'(a).$

(problem 2a) Find the derivative of the function $f(x) = x^2 - 3x + 5$ using the definition of the derivative, $f'(x) = \lim _{h \to 0} \frac {f(x+h) - f(x)}{h}.$

Step 1. Compute $f(x + h)$ :

$x^2 + 2xh + h^2 - 3x - 3h - 5$ $x^2 + 2xh + h^2 - 3x + 3h + 5$ $x^2 + 2xh + h^2 - 3x - 3h + 5$ $x^2 + 2xh + h^2 + 3x - 3h + 5$

Step 2. Compute the numerator of the difference quotient: $f(x+h) - f(x) =$

$2xh - h^2 - 3h$ $2xh + h^2 + 3h$ $2xh - h^2 + 3h$ $2xh + h^2 - 3h$

Step 3. Divide by $h$ to obtain the difference quotient: $\frac {f(x+h) - f(x)}{h} =$

$2x - h - 3$ $2x + h + 3$ $2x - h + 3$ $2x + h - 3$

Step 4. Take the limit as $h \to 0$ . The derivative is $f'(x) = \lim _{h \to 0} \frac {f(x+h) - f(x)}{h} = \; \answer {2x-3}.$

(problem 2b) Use the fact that the derivative of $f(x) = x^2 -3x + 5$ is $f'(x) = 2x-3$ , i.e., $\frac {d}{dx}\left ( x^2 - 3x + 5 \right ) = 2x-3,$ to find the slope of the tangent line to the graph of $y = x^2 - 3x + 5$ at the point where $x = 1$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(1) = (2x-3)\big |_{x=1}$

The slope is $\answer {-1}$

(problem 2c) Use the answer to the previous problem to find the equation of the tangent line to the graph of $y = x^2 - 3x + 5 \ \ \text {at} \ \ x=1.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(1, f(1))$

Solve for $y$

The equation of the tangent line is $y = \answer {-x+4}$

(problem 2d) Find the derivative of the function $f(x) = 3x^2 - 2x -4$ using the definition of the derivative, $f'(x) = \lim _{h \to 0} \frac {f(x+h) - f(x)}{h}.$

Step 1. Compute $f(x + h)$ :

$3x^2 + 6xh + 3h^2 + 2x + 2h - 4$ $3x^2 + 6xh + 3h^2 - 2x + 2h - 4$ $3x^2 + 6xh + 3h^2 - 2x - 2h - 4$ $3x^2 + 6xh + 3h^2 - 2x - 2h + 4$

Step 2. Compute the numerator of the difference quotient: $f(x+h) - f(x) =$

$6xh - 3h^2 - 2h$ $6xh + 3h^2 + 2h$ $6xh - 3h^2 + 2h$ $6xh + 3h^2 - 2h$

Step 3. Divide by $h$ to obtain the difference quotient: $\frac {f(x+h) - f(x)}{h} =$

$6x - 3h - 2$ $6x + 3h + 2$ $6x - 3h + 2$ $6x + 3h - 2$

Step 4. Take the limit as $h \to 0$ . The derivative is $f'(x) = \lim _{h \to 0} \frac {f(x+h) - f(x)}{h} = \; \answer {6x-2}.$

(problem 2e) Use the fact that the derivative of $3x^2 -2x - 4$ is $6x-2$ , i.e., $\frac {d}{dx}\left ( 3x^2 - 2x -4 \right ) = 6x-2,$ to find the slope of the tangent line to the graph of $y = 3x^2 - 2x - 4$ at the point where $x = 1$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(1)$

The slope is $\answer {4}$

(problem 2f) Use the answer to the previous problem to find the equation of the tangent line to the graph of $y = 3x^2 - 2x - 4 \ \ \text {at} \ \ x=1.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(1, f(1))$

Solve for $y$

The equation of the tangent line is $y = \answer {4x - 7}$

example 3 Compute the derivative of $f(x) = x^3$ .
We first compute $f(x+h)$ .

We have

$\begin{align*} f(x+h) &= (x+h)^3 \\ &= (x+h)(x+h)^2 \\ &= (x+h)(x^2 + 2xh + h^2)\\ &= x^3 + 2x^2h + xh^2 + x^2h + 2xh^2 + h^3\\ &= x^3 + 3x^2h + 3xh^2 + h^3. \end{align*}$ Now we use the definition: $\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3)- x^3}{h}\\[5pt] &= \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}\\[5pt] &= \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h}\\[5pt] &= \lim_{h \to 0} (3x^2 + 3xh + h^2) \\ &= 3x^2. \end{align*}$ To recap, $\frac {d}{dx}x^3 = 3x^2.$ Try to notice a pattern in the form of the derivatives of $x^2$ and $x^3$ .

(problem 3a) Use the fact that the derivative of $f(x) = x^3$ is $f'(x) = 3x^2$ , i.e., $\frac {d}{dx}\left ( x^3 \right ) = 3x^2,$ to find the slope of the tangent line to the graph of $y = x^3$ at the point where $x = 2$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(2)$

The slope is $\answer {12}$

(problem 3b) Use the answer to the previous problem to find the equation of the tangent line to the graph of $y = x^3 \ \ \text {at} \ \ x=2.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(2, f(2))$

Solve for $y$

The equation of the tangent line is $y = \answer {12x-16}$

example 4 Compute the derivative of $f(x) = \sqrt x$ .
Using the definition of the derivative, we have $\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{\sqrt{x+h}- \sqrt x}{h}\\[5pt] &= \lim_{h \to 0} \frac{\sqrt{x+h}- \sqrt x}{h} \cdot \frac{\sqrt{x+h}+ \sqrt x}{\sqrt{x+h}+ \sqrt x} \\[5pt] &= \lim_{h \to 0} \frac{(x+h) - (x)}{h(\sqrt{x+h}+ \sqrt x)}\\[5pt] &= \lim_{h \to 0} \frac{h}{h(\sqrt{x+h}+ \sqrt x)}\\[5pt] &= \lim_{h \to 0} \frac{1}{\sqrt{x+h}+ \sqrt x}\\[5pt] &= \frac{1}{2 \sqrt x}. \end{align*}$ To recap, $\frac {d}{dx}\sqrt {x} = \frac {1}{2\sqrt {x}},$ which in the notation of exponents, where $\sqrt x = x^{1/2}$ , we can write $\frac {d}{dx}x^{1/2} = \frac 12 x^{-1/2}.$

(problem 4a) Use the fact that the derivative of $f(x) = \sqrt {x}$ is $f'(x) = \frac {1}{2\sqrt {x}}$ , i.e., $\frac {d}{dx}\sqrt {x} = \frac {1}{2\sqrt {x}},$ to find the slope of the tangent line to the graph of $y = \sqrt {x}$ at the point where $x = 4$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(4)$

The slope is $\answer {1/4}$

(problem 4b) Use the answer to the previous problem to find the equation of the tangent line to the graph of $y = \sqrt {x} \ \ \text {at} \ \ x=4.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(4, f(4))$

Solve for $y$

The equation of the tangent line is $y = \answer {x/4 +1}$

example 5 Compute the derivative of $f(x) = \sqrt {2x+1}$ . Using the definition of the derivative, we have $\begin{align*} f(x+h) &= \sqrt{2(x+h)+1} \\ &= \sqrt{2x +2h+1}. \end{align*}$ Now for the derivative: $\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{\sqrt{2x+2h+1}- \sqrt{2x+1}}{h}\\[5pt] &= \lim_{h \to 0} \frac{\sqrt{2x+2h+1}- \sqrt{2x+1}}{h} \cdot \frac{\sqrt{2x+2h+1}+ \sqrt{2x+1}}{\sqrt{2x+2h+1}+ \sqrt{2x+1}} \\[5pt] &= \lim_{h \to 0} \frac{(2x+2h+1) - (2x+1)}{h(\sqrt{2x+2h+1}+ \sqrt{2x+1})}\\[5pt] &= \lim_{h \to 0} \frac{2h}{h\sqrt{2x+2h+1}+ \sqrt{2x+1}}\\[5pt] &= \lim_{h \to 0} \frac{2}{\sqrt{2x+2h+1}+ \sqrt{2x+1}}\\[5pt] &= \frac{2}{\sqrt{2x+1}+ \sqrt{2x+1}}\\[5pt] &= \frac{2}{2\sqrt{2x+1}}\\[5pt] &= \displaystyle{\frac{1}{\sqrt{2x+1}}}. \end{align*}$ In the notation of Leibniz, $\frac {d}{dx}\sqrt {2x+1}= \frac {1}{\sqrt {2x+1}}.$

(problem 5a) Use the fact that the derivative of $f(x) = \sqrt {2x+1}$ is $f'(x) = \frac {1}{\sqrt {2x+1}}$ , i.e., $\frac {d}{dx}\sqrt {2x+1} = \frac {1}{\sqrt {2x+1}},$ to find the slope of the tangent line to the graph of $y = \sqrt {2x+1}$ at the point where $x = 4$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(4)$

The slope is $\answer {1/3}$

(problem 5b) Use the answer to the previous problem to find the equation of the tangent line to the graph of $y = \sqrt {2x+1} \ \ \text {at} \ \ x=4.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(4, f(4))$

Solve for $y$

The equation of the tangent line is $y = \answer {x/3 + 5/3}$

(problem 5c) Find the derivative of the function $f(x) = \sqrt {2x + 5}$ using the definition of the derivative, $f'(x) = \lim _{h \to 0} \frac {f(x+h) - f(x)}{h}.$

Step 1. Compute $f(x + h)$ :

$\sqrt {2x + 2h + 5}$ $\sqrt {2x + h + 5}$ $\sqrt {2x + 5} + h$ $\sqrt {2x + 5} + 2h$

Step 2. Compute the numerator of the difference quotient, $f(x+h) - f(x)$ :

$\sqrt {2x + h + 5} - \sqrt {2x + 5}$ $\sqrt {2x + 5} + h- \sqrt {2x + 5}$ $\sqrt {2x + 5} + 2h - \sqrt {2x + 5}$ $\sqrt {2x + 2h + 5} - \sqrt {2x + 5}$

Step 3. Divide by $h$ to obtain the difference quotient: $\frac {f(x+h) - f(x)}{h} =$ (multiply by the conjugate radical and cancel $h$ )

$\frac {2}{\sqrt {2x + 2h + 5} + \sqrt {2x + 5}}$ $\frac {2}{\sqrt {2x + 2h + 5} - \sqrt {2x + 5}}$ $\frac {1}{\sqrt {2x + 2h + 5} + \sqrt {2x + 5}}$ $\frac {1}{\sqrt {2x + 2h + 5} - \sqrt {2x + 5}}$

Step 4. Take the limit as $h \to 0$ . The derivative is $f'(x) = \lim _{h \to 0} \frac {f(x+h) - f(x)}{h} = \; \answer {1/\sqrt {2x+5}}.$

(problem 5d) Use the fact that the derivative of $f(x) = \sqrt {2x + 5}$ is $f'(x) = 1/\sqrt {2x+5}$ , i.e., $\frac {d}{dx}\sqrt {2x+5} = \frac {1}{\sqrt {2x+5}},$ to find the slope of the tangent line to the graph of $f(x) = \sqrt {2x+5}$ at the point where $x = 2$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(1)$

The slope is $\answer {1/3}$

(problem 5e) Use the answer to the previous problem to find the equation of the tangent line to the graph of $y = \sqrt {2x+5} \ \ \text {at} \ \ x=2.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(2, f(2))$

Solve for $y$

The equation of the tangent line is $y = \answer {x/3 + 7/3}$

example 6 Compute the derivative of $f(x) = \displaystyle {\frac {1}{x}}$ .
Using the definition of the derivative, we have $\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{\frac{1}{x+h}- \frac{1}{x}}{h}\\[5pt] &= \lim_{h \to 0} \frac{\frac{(x) - (x+h)}{x(x+h)}}{h} \\[5pt] &= \lim_{h \to 0} \frac{-h}{x(x+h)}\cdot \frac{1}{h}\\[5pt] &= \lim_{h \to 0} \frac{-1}{x(x+h)} \\[5pt] &= -\frac{1}{x^2}. %\end{aligned}$ %\end{center} \end{align*}$ Using the differential operator, $d/dx$ , we can write: $\frac {d}{dx}\left ({\frac {1}{x}}\right ) = -\frac {1}{x^2}.$

(problem 6a) Use the fact that the derivative of $f(x) = \frac {1}{x}$ is $f'(x) = -\frac {1}{x^2}$ , i.e., $\frac {d}{dx}\left (\frac {1}{x}\right ) = -\frac {1}{x^2},$ to find the slope of the tangent line to the graph of $y = \frac {1}{x}$ at the point where $x = 2$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(2)$

The slope is $\answer {-1/4}$

(problem 6b) Use the answer to the previous problem to find the equation of the tangent line to the graph of $y = \frac {1}{x} \ \ \text {at} \ \ x=2.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(2, f(2))$

Solve for $y$

The equation of the tangent line is $y = \answer {-x/4 + 1}$

example 7 Compute the derivative of $f(x) = \displaystyle {\frac {2}{x+3}}$ .
Using the definition of the derivative, we have $\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{\frac{2}{x+h +3}- \frac{2}{x+3}}{h}\\[5pt] &= \lim_{h \to 0} \frac{\frac{2(x+3) - 2(x+h+3)}{(x+h+3)(x+3)}}{h} \\[5pt] &= \lim_{h \to 0} \frac{\frac{2x+6 - 2x-2h-6}{(x+h+3)(x+3)}}{h} \\[5pt] &= \lim_{h \to 0} \frac{-2h}{(x+h+3)(x+3)}\cdot \frac{1}{h}\\[5pt] &= \lim_{h \to 0} \frac{-2}{(x+h+3)(x+3)} \\[5pt] &= -\frac{2}{(x+3)^2}. \end{align*}$ To recap, the derivative of $\displaystyle {\frac {2}{x+3}}$ is $\displaystyle {-\frac {2}{(x+3)^2}}$ .

(problem 7a) Use the fact that the derivative of $f(x) = \frac {2}{x+3}$ is $f'(x) = -\frac {2}{(x+3)^2}$ , i.e., $\frac {d}{dx}\left (\frac {2}{x+3}\right ) = -\frac {2}{(x+3)^2},$ to find the slope of the tangent line to the graph of $y = \frac {2}{x+3}$ at the point where $x = -2$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(-2)$

The slope is $\answer {-2}$

(problem 7b) Use the answer to the previous problem to find the equation of the tangent line to the graph of $y = \frac {2}{x+3} \ \ \text {at} \ \ x=-2.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(-2, f(-2))$

Solve for $y$

The equation of the tangent line is $y = \answer {-2x-2}$

Video Lessons

Here are two detailed, lecture style videos on the derivative: