## Derivative Patterns

Read this section to learn about patterns of derivatives. Work through practice problems 1-8.

We have some general rules which apply to any elementary combination of differentiable functions, but in order to use the rules we still need to know the derivatives of each of the particular functions. Here we will add to the list of functions whose derivatives we know.

##### Derivatives of the Trigonometric Functions

We know the derivatives of the sine and cosine functions, and each of the other four trigonometric functions is just a ratio involving sines or cosines. Using the Quotient Rule, we can differentiate the rest of the trigonometric functions.

$\begin{array}{lll} \text { Theorem: } & \mathbf{D}(\tan (x)) \quad=\quad \sec ^{2}(x) & D(\sec (x))=\quad \sec (x) \tan (x) \\ & D(\cot (x)) \quad=-\csc ^{2}(x) & D(\csc (x))=-\csc (x) \cot (x) \end{array}$

Proof: From trigonometry we know $\tan (x)=\frac{\sin (x)}{\cos (x)}, \cot (x)=\frac{\cos (x)}{\sin (x)}, \sec (x)=\frac{1}{\cos (x)}$, and $\csc (x)=\frac{1}{\sin (x)}$ and we know $\mathbf{D}(\sin (x))=\cos (x)$ and $D(\cos (x))=-\sin (x) .$ Using the Quotient Rule,

\begin{align*} \begin{aligned} &\mathbf{D}(\tan (\mathrm{x}))=\mathbf{D}\left(\frac{\sin (\mathrm{x})}{\cos (\mathrm{x})}\right)=\frac{\cos (\mathrm{x}) \cdot \mathrm{D}(\sin (\mathrm{x}))-\sin (\mathrm{x}) \cdot \mathrm{D}(\cos (\mathrm{x}))}{(\cos (\mathrm{x}))^{2}} \\ &=\frac{\cos (\mathrm{x}) \cos (\mathbf{x})-\sin (\mathrm{x})\{-\sin (\mathbf{x})\}}{\cos ^{2}(\mathrm{x})}=\frac{\cos ^{2}(\mathrm{x})+\sin ^{2}(\mathrm{x})}{\cos ^{2}(\mathrm{x})}=\frac{1}{\cos ^{2}(\mathrm{x})}=\sec ^{2}(\mathbf{x}) \\ &=\frac{\cos (\mathrm{x})(\mathbf{0})-1\{-\sin (\mathbf{x})\}}{\cos ^{2}(\mathrm{x})}=\frac{\sin (\mathrm{x})}{\cos ^{2}(\mathrm{x})}=\frac{\sin (\mathrm{x})}{\cos (\mathrm{x})} \frac{1}{\cos (\mathrm{x})}=\tan (\mathbf{x}) \cdot \sec (\mathrm{x}) \end{aligned} \end{align*}

Instead of the Quotient Rule, we could have used the Power Rule to calculate $\mathbf{D}(\sec (x))=\mathbf{D}\left((\cos (\mathrm{x}))^{-1}\right)$.

Practice 3: Use the Quotient Rule on $f(x)=\cot (x)=\frac{\cos (x)}{\sin (x)}$ to prove that f $^{\prime}(x)=-\csc ^{2}(x)$.

Practice 4: Prove that $\mathbf{D}(\csc (x))=-\csc (x) \cdot \cot (x)$. The justification of this result is very similar to the justification for $\mathbf{D}(\sec (\mathrm{x}))$

Practice 5: Find (a) $\mathbf{D}\left(\mathrm{x}^{5} \cdot \tan (\mathrm{x})\right)$, (b) $\frac{\mathbf{d}}{\mathrm{dt}}\left(\frac{\sec (\mathrm{t})}{\mathrm{t}}\right)$ and (c) $\mathrm{D}(\sqrt{\cot (\mathrm{x})-\mathrm{x}})$.

##### Derivative of $e^x$

We can use graphs of exponential functions to estimate the slopes of their tangent lines or we can numerically approximate the slopes.

Example 3: Estimate the derivative of $\mathrm{f}(\mathrm{x})=2^{\mathrm{X}}$ at the point $\left(0,2^{0}\right)=(0,1)$ by approximating the slope of the line tangent to $\mathrm{f}(\mathrm{x})=2^{\mathrm{X}}$ at that point

Solution: We can get estimates from the graph of $\mathrm{f}(\mathrm{x})=2^{\mathrm{X}}$ by carefully graphing $\mathrm{f}(\mathrm{x})=2^{\mathrm{X}}$ for small values of $\mathrm{x}$, sketching secant lines, and then measuring the slopes of the secant lines (Fig. 1).

We can also find the slope numerically by using the definition of the derivative,

$\mathrm{f}^{\prime}(0) \equiv \lim\limits_{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim\limits_{h \rightarrow 0} \frac{2^{0+h}-2^{0}}{h}=\lim\limits_{h \rightarrow 0} \frac{2^{h}-1}{h}$, and evaluating $\frac{2^{\mathrm{h}}-1}{\mathrm{~h}}$ for some very small values of $\mathrm{h}$.

$h$ $\frac{2^{\mathrm{h}}-1}{\mathrm{~h}}$ $\frac{3^{\mathrm{h}}-1}{\mathrm{~h}}$ $\frac{e^{\mathrm{h}}-1}{\mathrm{~h}}$
0.1 0.717734625
-0.I 0.669670084
0.01 0.69555
-0.01 0.690750451
0.001 0.6933874
-0.001 0.69290695

$\begin{gathered} \downarrow \\ 0 \end{gathered}$

$\begin{gathered} \downarrow \\ \approx 0.693 \end{gathered}$

$\begin{gathered} \downarrow \\ \approx 1.099 \end{gathered}$

$\begin{gathered} \downarrow \\ 1 \end{gathered}$

From the table we can see that $f ' (0) ≈ .693$.

Practice 6: Fill in the table for $\frac{3^{\mathrm{h}}-1}{\mathrm{~h}}$, and show that the slope of the line tangent to $\mathrm{g}(\mathrm{x})=3^{\mathrm{x}}$ at $(0,1)$ is approximately $1.099$. (Fig. 2)

At $(0,1)$, the slope of the tangent to $y=2^{x}$ is less than 1, and the slope of the tangent to $\mathrm{y}=3^{\mathrm{x}}$ is slightly greater than 1. (Fig. 3) There is a number, denoted $\mathbf{e}$, between 2 and 3 so that the slope of the tangent to $\mathrm{y}=\mathrm{e}^{\mathrm{x}}$ is exactly 1: $\lim\limits_{h \rightarrow 0} \frac{e^{h}-1}{h}=1 .$ The number $\mathrm{e} \approx 2.71828182845904$.

$e$ is irrational and is very important and common in calculus and applications.

Once we grant that $\lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=1$, it is relatively straightforward to calculate $\mathbf{D}\left(\mathrm{e}^{\mathrm{x}}\right)$.

Theorem: $D(e^X) = e^X$

Proof:

\begin{align*} \begin{aligned} \mathbf{D}\left(\mathrm{e}^{\mathrm{x}}\right) & \equiv \lim\limits_{h \rightarrow 0} \frac{e^{x+h}-e^{x}}{h}=\lim\limits_{h \rightarrow 0} \frac{e^{x} \cdot e^{h}-e^{x}}{h} \\ &=\lim\limits_{h \rightarrow 0}\left(e^{x}\right) \cdot\left(\frac{e^{h}-1}{h}\right) \\ &=\lim\limits_{h \rightarrow 0}\left(e^{x}\right) \cdot \lim\limits_{h \rightarrow 0}\left(\frac{e^{h}-1}{h}\right)=\left(\mathrm{e}^{\mathrm{x}}\right)(1)=\mathbf{e}^{\mathbf{x}} . \end{aligned} \end{align*}

The function $f(x)=e^{x}$ is its own derivative: $f^{\prime}(x)=f(x)$. The height of $f(x)=e^{x}$ at any point and the slope of the tangent to $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$ at that point are the same: as the graph gets higher, its slope gets steeper.

Example 4: Find (a) $\frac{\mathbf{d}}{d t}\left(t \cdot e^{t}\right)$, (b) $D\left(e^{x} / \sin (x)\right)$ and (c) $D\left(e^{5 x}\right)=D\left(\left(e^{x}\right)^{5}\right)$

Solution: (a) Using the Product Rule with $\mathrm{f}(\mathrm{t})=\mathrm{t}$ and $\mathrm{g}(\mathrm{t})=\mathrm{e}^{\mathrm{t}}$,

\begin{align*} \frac{\mathbf{d}}{\mathbf{d t}}\left(t \cdot e^{t}\right)=t \cdot \mathbf{D}\left(\mathrm{e}^{\mathrm{t}}\right)+\mathrm{e}^{\mathrm{t}} \cdot \mathbf{D}(\mathrm{t})=\mathrm{t} \cdot \mathbf{e}^{\mathbf{t}}+\mathrm{e}^{\mathrm{t}} \cdot(\mathbf{1})=\mathrm{t} \cdot \mathrm{e}^{\mathrm{t}}+\mathrm{e}^{\mathrm{t}}=(\mathrm{t}+1) \mathrm{e}^{\mathrm{t}} \end{align*}

(b) Using the Quotient Rule with $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$ and $\mathrm{g}(\mathrm{x})=\sin (\mathrm{x})$,

\begin{align*} \mathbf{D}\left(\frac{\mathrm{e}^{\mathrm{x}}}{\sin (\mathrm{x})}\right)=\frac{\sin (\mathrm{x}) \mathbf{D}\left(\mathrm{e}^{\mathrm{x}}\right)-\mathrm{e}^{\mathrm{x}} \mathbf{D}(\sin (\mathrm{x}))}{\sin ^{2}(\mathrm{x})}=\frac{\sin (\mathrm{x}) \mathrm{e}^{\mathbf{x}}-\mathrm{e}^{\mathrm{x}} \cos (\mathbf{x})}{\sin ^{2}(\mathrm{x})} \end{align*}

(c) Using the Power Rule for Functions with $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$ and $\mathrm{n}=5$,

\begin{align*} D\left(\left(e^{x}\right)^{5}\right)=5\left(e^{x}\right)^{4} \cdot D\left(e^{x}\right)=5\left(e^{x}\right)^{4} \cdot e^{x}=5 e^{4 x} e^{x}=5 e^{5 x} \end{align*}

Practice 7: Find (a) $\mathrm{D}\left(\mathrm{x}^{3} \mathrm{e}^{\mathrm{x}}\right)$ and (b) $\mathrm{D}\left(\left(\mathrm{e}^{\mathrm{x}}\right)^{3}\right)$.