Textbook Problem

State each differentiation rule both in symbols and in words.

(a) The Power Rule

(b) The Constant Multiple Rule

(c) The Sum Rule

(d) The Difference Rule

(e) The Product Rule

(f) The Quotient Rule

(g) The Chain Rule

To determine

To state: The power rule.

Explanation of Solution

Statement:

If n is any real number, then $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$.

Example:

Differentiate, $f\left(x\right)=x^2$.

Obtain the derivative.

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(f\left(x\right)\right)\\ =\frac{d}{dx}\left(x^2\right)\\ =2x^{2-1}\left(\text{by power rule}\right)\\ =2x\end{array}$

To determine

To state: The constant multiple rule.

Explanation of Solution

Statement:

If c is a constant and f is a differentiable function, then

$\frac{d}{dx}\left(cf\left(x\right)\right)=c\frac{d}{dx}\left(f\left(x\right)\right)$

Example:

Differentiate, $f\left(x\right)=3x^2$.

Obtain the derivative of $f\left(x\right)$.

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(f\left(x\right)\right)\\ =\frac{d}{dx}\left(3x^2\right)\\ =3\frac{d}{dx}\left(x^2\right)\left(\text{by constant multiple rule}\right)\\ =6x\end{array}$

To determine

To state: The sum rule.

Explanation of Solution

Statement:

If $f_1\left(x\right)$ and $f_2\left(x\right)$ are both differentiable, then

$\frac{d}{dx}\left[f_1\left(x\right)+f_2\left(x\right)\right]=\frac{d}{dx}\left[f_1\left(x\right)\right]+\frac{d}{dx}\left[f_2\left(x\right)\right]$

Example:

Differentiate, $f\left(x\right)=x^2+2x$.

Obtain the derivative of $f\left(x\right)$.

$\begin{array}{c}\frac{d}{dx}\left[x^2+2x\right]=\frac{d}{dx}\left[x^2\right]+\frac{d}{dx}\left[2x\right]\left(\text{by sum rule}\right)\\ =2x^{2-1}+2\frac{d}{dx}\left[x\right]\\ =2x+2\left[1\right]\\ =2x+2\end{array}$

To determine

To state: The difference rule.

Explanation of Solution

Statement:

If $f_1\left(x\right)$ and $f_2\left(x\right)$ are both differentiable, then

$\frac{d}{dx}\left[f_1\left(x\right)-f_2\left(x\right)\right]=\frac{d}{dx}\left[f_1\left(x\right)\right]-\frac{d}{dx}\left[f_2\left(x\right)\right]$

Example:

Differentiate, $f\left(x\right)=x^2-2x$.

Obtain the derivative of $f\left(x\right)$.

$\begin{array}{c}\frac{d}{dx}\left[x^2-2x\right]=\frac{d}{dx}\left[x^2\right]-\frac{d}{dx}\left[2x\right]\left(\text{by difference rule}\right)\\ =2x^{2-1}-2\frac{d}{dx}\left[x\right]\\ =2x-2\left[1\right]\\ =2x-2\end{array}$

To determine

To state: The product rule.

Explanation of Solution

Statement:

If $f_1\left(x\right)$ and $f_2\left(x\right)$ are both differentiable, then

$\frac{d}{dx}\left[f_1\left(x\right)f_2\left(x\right)\right]=f_1\left(x\right)\frac{d}{dx}\left[f_2\left(x\right)\right]+f_2\left(x\right)\frac{d}{dx}\left[f_1\left(x\right)\right]$

Example:

Differentiate, $f\left(x\right)=x{e}^x$.

Obtain the derivative of $f\left(x\right)$.

$\begin{array}{c}\frac{d}{dx}\left[x{e}^x\right]=x\frac{d}{dx}\left[e^x\right]+e^x\frac{d}{dx}\left[x\right]\left(\text{by product rule}\right)\\ =x\left[e^x\right]+e^x\left[1\right]\\ =x{e}^x+e^x\end{array}$

To determine

To state: The quotient rule.

Explanation of Solution

Statement:

If $f_1\left(x\right)$ and $f_2\left(x\right)$ are both differentiable, then

$\frac{d}{dx}\left[\frac{f_1\left(x\right)}{f_2\left(x\right)}\right]=\frac{f_2\left(x\right)\frac{d}{dx}\left[f_1\left(x\right)\right]-f_1\left(x\right)\frac{d}{dx}\left[f_2\left(x\right)\right]}{{\left[f_2\left(x\right)\right]}^2}$

Example:

Differentiate of $f\left(x\right)=\frac{e^x}{x}$.

Obtain the derivative of $f\left(x\right)$.

$\begin{array}{c}\frac{d}{dx}\left[\frac{e^x}{x}\right]=\frac{x\frac{d}{dx}\left[e^x\right]-e^x\frac{d}{dx}\left[x\right]}{x^2}\left(\text{by quotient rule}\right)\\ =\frac{x\left[e^x\right]-e^x\left[1\right]}{x^2}\\ =\frac{x{e}^x-e^x}{x^2}\end{array}$

To determine

To state: The chain rule.

Explanation of Solution

Statement:

If g is differentiable at t and f is differentiable at $g\left(t\right)$, then the composite function $F=f\circ g$ defined by $F\left(t\right)=f\left(g\left(t\right)\right)$ is differentiable at x and $F^{\prime }$ is given by the product

$F^{\prime }\left(t\right)=f^{\prime }\left(g\left(t\right)\right)\cdot g^{\prime }\left(t\right)$

Example:

Differentiate $y={\log }_5\left(1+2x\right)$.

Obtain the derivative of $y$.

$\begin{array}{c}{y}^{\prime }=\frac{d}{dx}\left(y\right)\\ =\frac{d}{dx}\left({\log }_5\left(1+2x\right)\right)\end{array}$

Let $h\left(x\right)=1+2x$ and $f\left(u\right)={\log }_5\text{u where }u=h\left(x\right)$.

Apply the chain rule as shown in equation (1),

$y^{\prime }=f^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)$ (2)

The derivative $f^{\prime }\left(h\left(x\right)\right)$ is computed as follows,

$\begin{array}{c}{f}^{\prime }\left(h\left(x\right)\right)=f^{\prime }\left(u\right)\left[Qu=h\left(x\right)\right]\\ =\frac{d}{du}\left(f\left(u\right)\right)\\ =\frac{d}{du}\left({\log }_{5}u\right)\\ =\frac{1}{u\ln 5}\left(Q\frac{d}{dx}\left({\log }_{a}x\right)=\frac{1}{x\ln a}\right)\end{array}$

Substitute $u=1+2x$ in the above equation,

$f^{\prime }\left(h\left(x\right)\right)=\frac{1}{\left(1+2x\right)\ln 5}$

Thus, the derivative $f^{\prime }\left(h\left(x\right)\right)$ is $f^{\prime }\left(h\left(x\right)\right)=\frac{1}{\left(1+2x\right)\ln 5}$.

The derivative of $h\left(x\right)$ is computed as follows,

$\begin{array}{c}{h}^{\prime }\left(x\right)=\frac{d}{dx}\left(1+2x\right)\\ =\left(0+2\left(1\right)x^{1-1}\right)\\ =2\end{array}$

Thus, the derivative of $h\left(x\right)$ is $h^{\prime }\left(x\right)=2$.

Substitute $\frac{1}{\left(1+2x\right)\ln 5}$ for $f^{\prime }\left(h\left(x\right)\right)$ and $2$ for $h^{\prime }\left(x\right)$ in equation (2),

$\begin{array}{c}{y}^{\prime }=\frac{1}{\left(1+2x\right)\ln 5}\left(2\right)\\ =\frac{2}{\left(1+2x\right)\ln 5}\end{array}$

Therefore, the derivative of $y={\log }_5\left(1+2x\right)$ is $\underset{\_}{y^{\prime }=\frac{2}{\left(1+2x\right)\ln 5}}$.