Derivatives

Abigail Huettig

2 Derivatives

The derivative function $y'=f'(x)$ of $y=f(x)$ gives us the rate at which the dependent variable $y$ is changing in relation to the independent variable $x$ . Because the slope is the rate at which the dependent variable is changing in relation to the independent variable, the derivative can be thought of as slope. In fact, the derivative of a function $f$ at a point $x$ is the slope of the tangent line at $x$ . Consider the following function.

What we want to do is find the slope of the tangent line at $x=1$ in order to determine the slope of $f(x)$ at $x=1$ . We know that the slope of the line between two points is given as

$m=\frac{f(x_1)-f(x_2)}{x_1-x_2}.$

Let $x_1=1+\Delta x$ and $x_2=1$ . Now we have

$m=\frac{f(1+\Delta x)-f(1)}{\Delta x}.$

The problem is that this gives the slope of a secant line of $f(x)$ . So, how do we get the slope of the tangent line? We are going to approximate it with the slope of the secant line that is as close to being the tangent line as possible. We find this secant line by choosing $\Delta x$ to be as close to 0 as possible without $\Delta x$ equaling 0. Consider the following graph.

As $\Delta x$ gets closer to 0, the secant lines get closer to the tangent (purple) line. So, to find the slope of the tangent line, we need to find

$\lim_{\Delta x\rightarrow 0}\frac{f(1+\Delta x)-f(1)}{\Delta x}.$

Generalizing this result, we have the formal definition from Larson and Edwards [15].

Definition I.3
The derivative of $f$ at $x$ is

$f'(x)=\lim_{\Delta x\to 0} \displaystyle\frac{f(x+\Delta x)-f(x)}{\Delta x}$

provided the limit exists. For all $x$ for which this limit exists, $f'$ is a function of $x$ .

Additionally, the process of finding the derivative is known as differentiation, and the derivative of $y=f(x)$ with respect to $x$ is denoted as

$\frac{d}{dx}f(x)=f'(x)=y'=\frac{dy}{dx} .$

It is possible to use Definition I.3 to find derivatives by substituting $f(x)$ in and taking the limit as $\Delta x$ approaches 0. However, this process can become quite long and prone to error when dealing with complicated functions. In fact, finding derivatives using the definition is rarely used. Instead, the definition is used to prove derivative rules that make finding derivatives simpler and easier. Consider the following rules from Larson and Edwards [15].

The Constant Rule: $\displaystyle\frac{d}{dx}[c]=0$ where $c$ is a real number
The Power Rule: $\displaystyle\frac{d}{dx}[x^n]=nx^{n-1}$ where $n$ is a rational number
The Constant Multiple Rule: $\displaystyle\frac{d}{dx}[cf(x)]=cf'(x)$ where $c$ is a real number
The Sum and Difference Rules: $\displaystyle\frac{d}{dx}[f(x)\pm g(x)]=f'(x)\pm g'(x)$

Each of these rules can be proven by substituting into the formula given in the definition of a derivative, rewriting and simplifying, and then taking the limit. In the following pages, this will be demonstrated using the proofs for the product rule and quotient rule.

Let’s now put the above differentiation rules into practice by going through a problem I completed for Calculus I that comes from Larson and Edwards [15].

Example 3
Find the derivative of $f(x)=x^2+5-3x^{-2}$ .

Solution

$\begin{align*} \frac{d}{dx}[f(x)]&=\frac{d}{dx}[x^2+5-3x^{-2}]\\ &=\frac{d}{dx}[x^2]+\frac{d}{dx}[5]-\frac{d}{dx}[3x^{-2}] &&\text{Sum and Difference Rules}\\ &=\frac{d}{dx}[x^2]+\frac{d}{dx}[5]-3\left(\frac{d}{dx}[x^{-2}]\right) &&\text{Constant Multiple Rule}\\ &=2x+6x^{-3} &&\text{Constant and Power Rules}\\ \end{align*}$

Differentiating products and quotients is more complicated so we will look at the product and quotient rules separately, providing an example for each.

Product Rule

The following definition and proof of the product rule are from Larson and Edwards [15].

Theorem I.4

The product of two differentiable functions $f$ and $g$ is itself differentiable. Moreover, the derivative of $fg$ is the first function times the derivative of the second, plus the second function times the derivative of the first.

$\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

Proof.

$\begin{align*} \frac{d}{dx}[f(x)g(x)]&=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}\\ \text{}\\ &=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x+\Delta x)g(x)}{\Delta x}\;+\\ &\,\quad\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)g(x)-f(x)g(x)}{\Delta x} \\ \text{}\\ &=\lim_{\Delta x\rightarrow 0}f(x+\Delta x)\frac{g(x+\Delta x)-g(x)}{\Delta x}\;+\\ &\,\quad\lim_{\Delta x\rightarrow 0}g(x)\frac{f(x+\Delta x)-f(x)}{\Delta x}\\ \text{}\\ &=\lim_{\Delta x\rightarrow 0}f(x+\Delta x)\cdot \lim_{\Delta x\rightarrow 0}\frac{g(x+\Delta x)-g(x)}{\Delta x}\;+\\ &\,\quad\lim_{\Delta x\rightarrow 0}g(x)\cdot \lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\\ \text{}\\ &=f(x)g'(x)+g(x)f'(x) \end{align*}$

Now, let’s practice using a problem I completed for Calculus I that comes from Larson and Edwards [15].

Example 4
Find the derivative of $f(x)=(x^2+3)(x^2-4x)$ using the product rule.

Solution

$\begin{align*} \frac{d}{dx}[f(x)]&=\frac{d}{dx}[(x^2+3)(x^2-4x)]\\ &=(x^2+3)\left(\frac{d}{dx}[x^2-4x]\right)+(x^2-4x)\left(\frac{d}{dx}[x^2+3]\right) \\ &=(x^2+3)(2x-4)+(x^2-4x)(2x)\\ &=2x^3-4x^2+6x-12+2x^3-8x^2\\ &=4x^3-12x^2+6x-12 \end{align*}$

Therefore, the derivative function of $f(x)=(x^2+3)(x^2-4x)$ is

$f'(x)=4x^3-12x^2+6x-12.$

Interpreting this result, the slope of $f(x)$ at $x$ is $f'(x)$ . Equivalently, the rate at which $f(x)$ is changing at $x$ is $f'(x)$ .

Quotient Rule

The following definition and proof of the quotient rule are from Larson and Edwards as well [15].

Theorem I.5

The quotient $f/g$ of two differentiable functions $f$ and $g$ is itself differentiable at all values of $x$ for which $g(x)\neq 0.$ Moreover, the derivative of $f/g$ is given by the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$

Proof.

$\begin{align*} \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]&=\lim_{\Delta x\rightarrow 0}\frac{\displaystyle\left[\frac{f(x+\Delta x)}{g(x+\Delta x)}-\displaystyle\frac{f(x)}{g(x)}\right]}{\Delta x}\\ \text{}\\ &=\lim_{\Delta x\rightarrow 0}\frac{g(x)f(x+\Delta x)-f(x)g(x+\Delta x)}{\Delta x\cdot g(x)\cdot g(x+\Delta x)}\\ \text{}\\ &=\lim_{\Delta x\rightarrow 0}\frac{g(x)f(x+\Delta x)-f(x)g(x)+f(x)g(x)-f(x)g(x+\Delta x)}{\Delta x\cdot g(x)\cdot g(x+\Delta x)}\\ \text{}\\ &=\displaystyle\lim_{\Delta x\rightarrow 0}\displaystyle\frac{\left[\displaystyle\frac{g(x)[f(x+\Delta x)-f(x)]-f(x)[g(x+\Delta x)-g(x)]}{\Delta x}\right]}{g(x)\cdot g(x+\Delta x)}\\ \text{}\\ &=\frac{\displaystyle\lim_{\Delta x\rightarrow 0}\frac{g(x)[f(x+\Delta x)-f(x)]}{\Delta x}-\displaystyle\lim_{\Delta x\rightarrow 0}\frac{f(x)[g(x+\Delta x)-g(x)]}{\Delta x}}{\displaystyle\lim_{\Delta x\rightarrow 0}(g(x)\cdot g(x+\Delta x))}\\ \text{}\\ &=\frac{g(x)\left[\displaystyle\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\right]-f(x)\left[\displaystyle\lim_{\Delta x\rightarrow 0}\frac{g(x+\Delta x)-g(x)}{\Delta x}\right]}{\displaystyle\lim_{\Delta x\rightarrow 0}(g(x)\cdot g(x+\Delta x))}\\ \text{}\\ &=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2} \end{align*}$

Let’s now practice using a problem I completed for Calculus I that comes from Larson and Edwards [15].

Example 5
Find the derivative of $f(x)=\displaystyle\frac{x}{x^2+1}$ .

Solution

$\begin{align*} \frac{d}{dx}[f(x)]&=\frac{d}{dx}\left[\frac{x}{x^2+1}\right]\\ &=\frac{(x^2+1)\frac{d}{dx}[x]-(x)\frac{d}{dx}[x^2+1]}{(x^2+1)^2} &&\text{Quotient Rule}\\ &=\frac{(x^2+1)(1)-(x)(2x)}{(x^2+1)^2}\\ &=\frac{x^2+1-2x^2}{(x^2+1)^2}\\ &=\frac{-x^2+1}{(x^2+1)^2}\\ \end{align*}$

Therefore, the derivative function of $f(x)=\displaystyle\frac{x}{x^2+1}$ is

$f'(x)=\frac{-x^2+1}{(x^2+1)^2}.$

The next rule we will discuss is the chain rule which enables us to differentiate functions within functions.

Chain Rule

Consider the function $y=(2x+1)^2.$ How would we differentiate this? From the power rule, it would seem that

$\frac{dy}{dx}=2(2x+1)=4x+2.$

But is this actually correct? Let us expand $y$ to get $y=4x^2+4x+1.$ In this case,

$\frac{dy}{dx}=8x+4 \neq 4x+2.$

We cannot simply use the power rule because we do not have $x$ raised to a power but rather a function of $x$ raised to a power. We can think of $y$ as being $f(u)$ where $u=g(x)=2x+1$ . When we only used the power rule, we did not differentiate $f$ with respect to $x$ but rather with respect to $u$ . So, how do we get from $f'(u)$ to $f'(x)$ ? The solution to this problem is the chain rule, which is stated by Larson and Edwards as follows [15].

Theorem I.6

If $y=f(u)$ is a differentiable function of $u$ and $u=g(x)$ is a differentiable function of $x$ , then $y=f(g(x))$ is a differentiable function of $x$ and

$\frac{dy}{dx}=\frac{dy}{du}\left(\frac{du}{dx}\right)$

or, equivalently,

$\frac{d}{dx}[f(g(x))]=f'(g(x))g'(x).$

In other words, the solution is to multiply $f'(u)$ by $g'(x)$ . The theorem states it as

$\frac{dy}{du}\left(\frac{du}{dx}\right)=\frac{dy}{dx}$

because $du$ cancels. So, going back to our example, what is $\displaystyle\frac{dy}{dx}$ when

$y=f(g(x))=(2x+1)^2?$

We have $y=f(u)=u^2$ and $u=g(x)=2x+1$ . By the chain rule,

$\frac{dy}{dx}=\frac{dy}{du}\left(\frac{du}{dx}\right)=2u(2)=4(2x +1)=8x+4.$

Let’s now put the chain rule into practice with a more complicated function. This is a problem I completed for Calculus I and comes from Larson and Edwards [15].

Example 6
Find the derivative of $f(x)=\sqrt[3]{6x^2+1}$ .

Solution
Let $h(g(x))=(g(x))^{\frac{1}{3}}$ and $g(x)=6x^2+1.$

$\begin{align*} \frac{d}{dx}f(x)&=\frac{d}{dx}(h(g(x))\\ &=h'g(x)\cdot g'(x) &&\text{Chain Rule}\\ &=\frac{1}{3}(g(x))^{-\frac{2}{3}}\cdot (12x) \\ &=\frac{4x}{\sqrt[3]{(g(x))^2}}\\ &=\frac{4x}{\sqrt[3]{(6x^2+1)^2}}\\ \end{align*}$

Before moving on to integrals, we are going to discuss a special type of differentiation, known as implicit differentiation. As the name suggests, implicit differentiation is the process we use to differentiate implicit functions.

Implicit Differentiation

There are two forms that a function can take, implicit and explicit. An explicit function is one where one variable is clearly a function of another variable or variables. This type of function is what we have been dealing with so far. For example,

$y=\sqrt{1-x^2}$

is an explicit function because $y$ is clearly a function of $x$ . On the other hand,

$x^2+y^2=1; x,y\geq 0$

is an implicit function because it is not clear whether $x$ is a function of $y$ or vice versa.

For this particular equation, we restrict both $x$ and $y$ to be greater than or equal to 0 because this equation may not be a function otherwise. If $x$ is a function of $y$ , we can solve for $x$ to get

$x=\pm\sqrt{1-y^2}.$

Because the square root can be positive or negative, we have

$x=+\sqrt{1-y^2}\text{ and } x=-\sqrt{1-y^2}.$

This is not a function because $f(y)=\pm x$ . Thus, we must restrict $x$ to being nonnegative so that it is a function. This works in a similar manner for if $y$ is a function of $x$ . Because we do not know whether $x$ is a function of $y$ or vice versa, we must restrict both $x$ and $y$ to being nonnegative so that we are guaranteed $x^2+y^2=1$ is an implicit function. For a more detailed discussion of functions, see Part III: Chapter 15.

Sometimes, the implicit form does not cause too much trouble when differentiating. This is when we can easily solve for either one of the variables to be in terms of the other variable. Then, we can differentiate like normal. However, it is not always easy to solve for a variable. Take, for example,

$x^3+7y^5+8y^4=16.$

It is not clear how one might solve for $y$ in order to find $\frac{dy}{dx}$ . This is where the process of implicit differentiation comes in. The steps of implicit differentiation are outlined by Larson and Edwards as follows [15].

Differentiate both sides of the equation.
Move all the terms with $\frac{dy}{dx}$ to one side of the equation and all other terms to the other side.
Factor out $\frac{dy}{dx}$ .
Solve for $\frac{dy}{dx}$ .

Because we are solving for $\frac{dy}{dx}$ , it is assumed that $y$ is a function of $x$ . That being said, it is easier to understand implicit differentiation if we let $y=f(x)$ before beginning these steps. Let’s now practice with a problem I completed for Calculus I from Larson and Edwards [15].

Example 7
Find $\displaystyle\frac{dy}{dx}$ given that $x^3y^3-y=x$ .

Solution
Let $y=f(x)$ .

$\begin{align*} \frac{d}{dx}[x^3y^3-y]&=\frac{d}{dx}[x] &&\text{Differentiate w.r.t. $x$}\\ \frac{d}{dx}[x^3(f(x))^3-f(x)]&=\frac{d}{dx}[x] \\ x^3\frac{d}{dx}[(f(x))^3]+(f(x))^3\frac{d}{dx}[x^3]-\frac{d}{dx}[f(x)]&=\frac{d}{dx}[x] \\ x^3\left(3(f(x))^2 \frac{d}{dx}f(x)\right)+(f(x))^3(3x^2)-\frac{d}{dx}f(x)&=1 \\ x^3\left(3y^2 \frac{dy}{dx}\right)+y^3(3x^2)-\frac{dy}{dx}&=1 \\ 3x^3y^2\frac{dy}{dx}-\frac{dy}{dx}&=-3x^2y^3+1 &&\text{Move terms}\\ \frac{dy}{dx}(3x^3y^2-1)&=-3x^2y^3+1 &&\text{Factor out $\frac{dy}{dx}$}\\ \frac{dy}{dx}&=\frac{-3x^2y^3+1}{3x^3y^2-1} &&\text{Solve for $\frac{dy}{dx}$}\\ \end{align*}$

Therefore, the derivative of $y$ with respect to $x$ where $x^3y^3-y=x$ is

$\frac{dy}{dx}=\frac{-3x^2y^3+1}{3x^3y^2-1}.$

This means that if we could solve $x^3y^3-y=x$ for $y$ , we would have that $y$ is a function of $x$ , $y=f(x)$ , and the slope of $f(x)$ at $x$ would be

$f'(x)=\frac{-3x^2[f(x)]^3+1}{3x^3[f(x)]^2-1}.$

We are now ready to begin discussing integrals which are like the other side of the derivative coin and are even sometimes referred to as antiderivatives.

2 Derivatives

Product Rule

Quotient Rule

Chain Rule

Implicit Differentiation

License

Share This Book