3.1 Rules of Differentiation

Example 1

Find the derivative of $y=f(x)=mx+b$

This is a linear function, so its graph is its own tangent line! The slope of the tangent line, the derivative, is the slope of the line: $$f^{\prime }(x)=m$$

Example 2

Find the derivative of $f(x)=135$.

Think about this one graphically, too. The graph of f(x) is a horizontal line. So its slope is zero: $$f^{\prime }(x)=0$$

Example 3

Find the derivative of $f(x)=x^2$.

Recall the formal definition of the derivative: $$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}.$$

Using our function $f(x)=x^2$, $f(x+h)=(x+h{)}^2=x^2+2xh+h^2$.

Then
$$\begin{array}{rl}{f}^{\prime }(x)=& \underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ =& \underset{h\to 0}{lim}\frac{x^2+2xh+h^2-x^2}{h}\\ =& \underset{h\to 0}{lim}\frac{2xh+h^2}{h}\\ =& \underset{h\to 0}{lim}\frac{h(2x+h)}{h}\\ =& \underset{h\to 0}{lim}(2x+h)\\ =& 2x\end{array}$$

From all that, we find that $f^{\prime }(x)=2x$.

Example 4

Find the derivative of $g(x)=4x^3$

Using the power rule, we know that if $f(x)=x^3$, then $f^{\prime }(x)=3x^2$. Notice that $g$ is 4 times the function $f$. Think about what this change means to the graph of $g$ – it’s now 4 times as tall as the graph of $f$. If we find the slope of a secant line, it will be $\frac{\mathrm{\Delta }g}{\mathrm{\Delta }x}=\frac{4\mathrm{\Delta }f}{\mathrm{\Delta }x}=4\frac{\mathrm{\Delta }f}{\mathrm{\Delta }x}$; each slope will be 4 times the slope of the secant line on the $f$ graph. This property will hold for the slopes of tangent lines, too: $$\frac{d}{dx}\left(4x^3\right)=4\frac{d}{dx}\left(x^3\right)=4\cdot 3x^2=12x^2.$$

Derivative Rules: Building Blocks

In what follows, $f$ and $g$ are differentiable functions of $x$.

Constant Multiple Rule

$$\frac{d}{dx}\left(kf\right)=k{f}^{\prime }$$

Sum and Difference Rule

$$\frac{d}{dx}(f\pm g)=f^{\prime }\pm g^{\prime }$$

Power Rule

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

Special cases:
$$\frac{d}{dx}\left(k\right)=0\text{Because}k=k{x}^0.$$
$$\frac{d}{dx}\left(x\right)=1\text{Because}x=x^1.$$

Exponential Functions

$$\frac{d}{dx}\left(e^x\right)=e^x$$ $$\frac{d}{dx}\left(a^x\right)=\ln \left(a\right)a^x$$

Natural Logarithm

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

Example 5

Find the derivative of $p(x)=17x^{10}+13x^8-1.8x+1003$.

$\frac{d}{dx}(17x^{10}+13x^8-1.8x+1003)=$
$$\begin{array}{rl}=& \frac{d}{dx}\left(17x^{10}\right)+\frac{d}{dx}\left(13x^8\right)-\frac{d}{dx}\left(1.8x\right)+\frac{d}{dx}\left(1003\right)\\ =& 17\frac{d}{dx}\left(x^{10}\right)+13\frac{d}{dx}\left(x^8\right)-1.8\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(1003\right)\\ =& 17\left(10x^9\right)+13\left(8x^7\right)-1.8\left(1\right)+0\\ =& 170x^9+104x^7-1.8\end{array}$$

Example 6

Find $\frac{d}{dx}(17x^2-33x+12)$.

Writing out the rules, we’d write $$\frac{d}{dx}(17x^2-33x+12)=17(2x)-33(1)+0=34x-33.$$

Once you’re familiar with the rules, you can, in your head, multiply the 2 times the 17 and the 33 times 1, and just write $$\frac{d}{dx}(17x^2-33x+12)=34x-33.$$

Example 7

Find the derivative of $y=3\sqrt{t}-\frac{4}{t^4}+5e^t$.

The first step is translate into exponents: $$y=3\sqrt{t}-\frac{4}{t^4}+5e^t=3t^{1/2}-4t^{-4}+5e^t$$

Now you can take the derivative:
$$\begin{array}{rl}\frac{d}{dt}(3t^{1/2}-4t^{-4}+5e^t)=& 3\left(\frac{1}{2}{t}^{-1/2}\right)-4(-4t^{-5})+5\left(e^t\right)\\ =& \frac{3}{2}{t}^{-1/2}+16t^{-5}+5e^t\end{array}$$

If there is a reason to, you can rewrite the answer with radicals and positive exponents: $$y^{\prime }=\frac{3}{2}{t}^{-1/2}+16t^{-5}+5e^t=\frac{3}{2\sqrt{t}}+\frac{16}{t^5}+5e^t$$

Example 8

Find the equation of the line tangent to $g(t)=10-t^2$ when $t=2$.

The slope of the tangent line is the value of the derivative. We can compute $g^{\prime }(t)=-2t$. To find the slope of the tangent line when $t=2$, evaluate the derivative at that point. The slope of the tangent line is -4.

To find the equation of the tangent line, we also need a point on the tangent line. Since the tangent line touches the original function at $t=2$, we can find the point by evaluating the original function: $g(2)=10-2^2=6$. The tangent line must pass through the point (2, 6).

Using the point-slope equation of a line, the tangent line will have equation $y-6=-4(t-2)$. Simplifying to slope-intercept form, the equation is $y=-4t+14$.

Graphing, we can verify this line is indeed tangent to the curve:

Long description: A tangent line is shown passing through the point (2, 6). The x-axis extends from -1 to 4 and the y-axis extends from -1 to 11.

Example 9

In a memory experiment, a researcher asks the subject to memorize as many words from a list as possible in 10 seconds. Recall is tested, then the subject is given 10 more seconds to study, and so on. Suppose the number of words remembered after $t$ seconds of studying could be modeled by $W(t)=4t^{2/5}$. Find and interpret $W^{\prime }(20)$.

$W^{\prime }(t)=4\cdot \frac{2}{5}{t}^{-3/5}=\frac{8}{5}{t}^{-3/5}$, so $W^{\prime }(20)=\frac{8}{5}(20{)}^{-3/5}\approx 0.2652$.

Since $W$ is measured in words, and $t$ is in seconds, $W^{\prime }$ has units words per second. $W^{\prime }(20)\approx 0.2652$ means that after 20 seconds of studying, the subject is learning about 0.27 more words for each additional second of studying.