Rules of Differentiation - Mathematics.lu

In this page, we develop the remaining rules to find derivatives. Towards the end, we make a summary to have all the rules together and this will enable us to find any derivative by referring to these rules carefully.

We saw that give two functions $f$ and $g$ , the derivative of their sum $(f+g)'$ is the same as the sum of the derivatives $f'+g'$ . We would expect the product and quotient to be the same, but this is not the case. Remember that the derivative is defined as a limit:

$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

If we want to find the derivative of $f\cdot g$ , then we would have $(f\cdot g)(x+h)$ and not $f(x+h)\cdot g(x+h)$ . This already creates a big difference in the definition of our derivative, so we need new rules for product, quotient, and composition of functions.

Table of Contents

Product Rule

Claim (The Product Rule): If $f(x)$ and $g(x)$ are both differentiable real-valued functions, then the derivative of their product function is equal to the derivative of the first function multiplied by the second function plus the first function multiplied by the derivative of the second function.

$(f(x) \cdot g(x))' = f'(x)g(x) + f(x)g'(x)$

Let us see how we use this rule with a simple example:

Example: Use the product rule to determine the derivative of the function $y = x^3 \cdot \sin(x)$ .

Before we move on, one final example showing the product of three functions, and using the product rule twice:

Example: Determine the derivative of the triple product function $y = x^2 \cdot e^x \cdot \sin(x)$ by applying the standard product rule twice sequentially.

A general “product rule” with three functions $f,g,$ and $h$ would be:

$\Big(f(x)\cdot g(x)\cdot h(x)\Big)'=f'(x)\cdot (g(x)\cdot h(x))+g'(x)\cdot (f(x)\cdot h(x))+h'(x)\cdot(g(x)\cdot f(x))$

Quotient Rule

Like the product rule, the quotient rule has its own formula, and is simply a matter of copy-pasting the elements into your formula and working in steps.

Claim (The Quotient Rule): If $f(x)$ and $g(x)$ are both differentiable functions and $g(x) \neq 0$ , then the derivative of their quotient function is given by:

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

Example: Use the quotient rule to determine the derivative of the function $y = \frac{x^3}{\sin(x)}$

Chain Rule

From experience, this is the rule that students struggle with the most. The idea is the following: we know how to find the derivative of $e^x$ , for example. But how do we find the derivative of $e^{x^2+4x-3}$ ? Here, the exponent is not simply $x$ , but some function $f(x)$ . This is an example of a composition of functions. The chain rule tells us to differentiate the outer function “as if $f(x)$ were just an $x$ ,” and then compensate for the fact that $f(x)$ itself depends on $x$ by multiplying the result by $f'(x)$ .

Claim (The Chain Rule): If $g(x)$ is a differentiable function at $x$ and $f(u)$ is a differentiable function at $u = g(x)$ , then the composite function $y = f(g(x))$ is differentiable at $x$ , and its derivative is equal to the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function.

$\frac{\mathrm{d}}{\mathrm{d}x}[f(g(x))] = f'(g(x)) \cdot g'(x)$

There is no need to understand the proof above. It is not entirely formal, but it is a neat trick that treats derivatives somewhat like fractions and takes advantage of the notation $\frac{dy}{dx}$ . While $\frac{dy}{dx}$ is not literally a fraction in the usual sense, this way of thinking provides useful intuition for the chain rule.

Example: Use the chain rule to determine the derivative of the composite function $y = \sin(x^3 + 5x)$ .

The next example is meant mostly to cover harder situations and show that as long as you work in terms, you can use the chain rule to find any derivative.

Example: Use the chain rule twice to determine the derivative of the triple composite function $y = \ln(\sin(x^2 + 1))$ .

Summary

Combining the claims above with the ones from yesterday, here are all the rules necessary to find derivatives.

Claim (Summary of Fundamental Differentiation Rules): Let $f(x)$ and $g(x)$ be differentiable real-valued functions, and let $k \in \mathbb{R}$ be any constant scaling parameter. The core structural operations of calculus satisfy the following five analytical rules:

1. Sum and Difference Rule:
$(f(x) \pm g(x))' = f'(x) \pm g'(x)$
2. Constant Multiple Rule:
$(k \cdot f(x))' = k \cdot f'(x)$
3. Product Rule:
$(f(x) \cdot g(x))' = f'(x)g(x) + f(x)g'(x)$
4. Quotient Rule (where $g(x) \neq 0$ ):
$\left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$
5. Chain Rule:
$\frac{\mathrm{d}}{\mathrm{d}x}[f(g(x))] = f'(g(x)) \cdot g'(x)$

Exercises

Calculate the derivative with respect to $x$ for each of the following functions. Apply the product rule formula $(f \cdot g)' = f'g + fg'$ .

$y = x^5 \cdot e^x$

$y = \cos(x) \cdot \ln(x)$

$y = \sqrt{x} \cdot \sin(x)$

$y = (3x^2 - 5x) \cdot e^x$

Calculate the derivative with respect to $x$ for each of the following three functions using the product rule.

$y = x^3 \cdot \ln(x) \cdot \cos(x)$

$y = \sqrt{x} \cdot e^x \cdot \tan(x)$

$y = x \cdot \sin(x) \cdot \ln(x)$

Calculate the derivative with respect to $x$ for each of the following functions. Apply the quotient rule and simplify your expressions completely.

$y = \frac{e^x}{x^4}$

$y = \frac{\ln(x)}{\cos(x)}$

$y = \frac{\sin(x)}{\sqrt{x}}$

$y = \frac{5x - 2}{x^2 + 3}$

Use the fact that: $\tan(x) = \frac{\sin(x)}{\cos(x)}$ and the quotient rule to show that the derivative of $\tan(x)$ is $\frac{1}{\cos^2(x)}$ .

Calculate the derivative with respect to $x$ for each of the following four composite functions. Clearly identify the inner and outer operations before using the chain rule.

$y = (4x^2 - 3x + 1)^5$

$y = e^{\cos(x)}$

$y = \ln(5x^3 + 2)$

$y = \sqrt{\sin(x)}$

Calculate the derivative with respect to $x$ for each of the following three triple composite functions. You will need to apply the chain rule twice sequentially, working from the outermost layer to the innermost core layer.

$y = e^{\sin(3x^4)}$

$y = \sin^4(\ln(x)) = (\sin(\ln(x)))^4$

$y = \ln\left(\cos\left(e^x\right)\right)$

Calculate the derivative with respect to $x$ for each of the following three functions. Each problem requires you to nest the chain rule inside a product or quotient rule framework.

$y = x^4 \cdot e^{\sin(x)}$

$y = \frac{\ln(3x^2 + 1)}{x^2}$

$y = \cos^3(x) \cdot \sqrt{x}$