# Techniques to Evaluate Derivatives

## Basic Techniques

Last time, we talked about some of the different properties of derivatives and the limit definition of the derivative. As another example, lets try evaluating the derivative of $f(x)=x^5$. Using the limit definition and also expanding out the binomial term, we have

$\frac{df}{dx}=\lim_{h\rightarrow 0}\frac{(x+h)^5-x^5}{h}=\lim_{h\rightarrow 0}\frac{5x^4h+10x^3h^2+10x^2h^3+5xh^4+h^5}{h}$

Simplifying gives

$\frac{df}{dx}=\lim_{h\rightarrow 0}(5x^4+10x^3h+10x^2h^2+5xh^3+h^4)=5x^4$

As we can see, this process is doable, but tedious and algebraically hairy. Let’s develop a set of rules to make finding derivatives a little easier:

1. If $f(x)=\text{ constant}$, then $f’(x)=0$.
2. If $f(x)=mx+b$, then $f’(x)=m$.
3. Power Rule: If $f(x)=x^n$, then $f’(x)=nx^{n-1}$. Note that $n$ does not have to be an integer; it can be any real number!
4. Linearity: If $h(x)=\alpha f(x)\pm \beta g(x)$ for constants $\alpha, \beta$, then $h’(x)=\alpha f’(x)\pm \beta g’(x)$.
5. Product Rule: If $h(x)=f(x)g(x)$, then $h(x)=f’(x)g(x)+g’(x)f(x)$.
6. Quotient Rule: If $h(x)=f(x)/g(x)$, then
$h(x)=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}$

These are the basic rules. Of course, there are more advanced techniques possible to evaluate derivatives, but we’ll introduce them in the future. Let’s try a few examples:

### Example 1

Let’s evaluate the derivative of $f(x)=(x+1)^2$ using the power rule and linearity. We can expand $f(x)$ as $f(x)=x^2+2x+1$. The derivative of $x^2$ is $2x$, the derivative of $x$ is $1$, and the derivative of $1$ is $0$ using the power rule. Therefore, we have

$\frac{d}{dx}[(x+1)^2]=\frac{d}{dx}[x^2]+2\frac{d}{dx}[x]+\frac{d}{dx}[1]=2x+2(1)+0=2x+2$

### Example 2

Let’s evaluate the derivative of $f(x)=x^2=x\cdot x$ using the product rule. Here $f(x)=x$ (meaning $f’(x)=1$) and $g(x)=x$ (meaning $g’(x)=1$). Plugging this into the formula from above, we find

$\frac{d}{dx}[x^2]=(1)(x)+(1)(x)=2x$

We can also evaluate this derivative using the power rule, where $n=2$:

$\frac{d}{dx}[x^2]=2x^{2-1}=2x$

Notice how we get the same result using either method of finding the derivative.

### Example 3

Let’s evaluate the derivative of $f(x)=1/x$ using the quotient rule. Here $f(x)=1$ (meaning $f’(x)=0$) and $g(x)=x$ (meaning $g’(x)=1$). Plugging this into the formula from above, we find

$\frac{d}{dx}\left[\frac{1}{x}\right]=\frac{(0)(x)-(1)(1)}{x^2}=\frac{-1}{x^2}=-x^{-2}$

We can also evaluate this derivative using the power rule, where $n=-1$:

$\frac{d}{dx}\left[\frac{1}{x}\right]=\frac{d}{dx}\left[x^{-1}\right]=-1x^{-1-1}=-x^{-2}$

Notice how we get the same result using either method of finding the derivative.

## Exercises

### Problem 1

Evaluate the derivatives of the following functions:

1. $f(x)=4x^5-5x^4$
2. $f(x)=\frac{x}{1+x^2}$
3. $f(x)=\frac{x^2-1}{x}$
4. $f(x)=(3x^2)\sqrt{x}$
5. $f(x)=2x-\frac{4}{\sqrt{x}}$
6. $f(x)=(x^2+3)(x^3+4)$
7. $f(x)=\frac{\frac{1}{x}+\frac{1}{x^2}}{x-1}$ Hint: It may be easier to rewrite this function first.

### Problem 2

Assume $f$ and $g$ are differentiable functions such that $f(2)=3$, $f’(2)=-1$, $f’(3)=7$, $g(2)=-5$, and $g’(2)=2$. Find the numerical value of the following expressions:

1. $(g-f)’(2)$
2. $(fg)’(2)$
3. $(\frac{f}{g})’(2)$
4. $(5f+3g)’(2)$