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. $(5f+3g)’(2)$