We now know how to differentiate basic algebraic and trigonometric functions but how we differentiate the compositions, products and quotients of these functions?
Chain rule
Given the composition of two functions, $$ \begin{equation}\begin{aligned} y=f(u)\text{ where } u=g(x)\\ \end{aligned}\end{equation} $$
The derivative will be $$ \begin{equation}\begin{aligned} \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\\ \end{aligned}\end{equation} $$
Example 1
For example, consider the function $$ \begin{equation}\begin{aligned} y=\sin(x+1)\\ \end{aligned}\end{equation} $$ This function can be achieved by substituting $x+1$ into $\sin{x}$. Thus, it is a composition of two functions. We can let the inner function $x+1$ be $u$: $$ \begin{equation}\begin{aligned} u&=x+1\\ \therefore y&=\sin{u}\\ \end{aligned}\end{equation} $$
Since we have $y$ in terms of $u$, we find the derivative $$ \begin{equation}\begin{aligned} \frac{dy}{du}&=\cos{u}\\ \end{aligned}\end{equation} $$ From the substitution, we found $u$ in terms of $x$ so we can have the derivative $$ \begin{equation}\begin{aligned} \frac{du}{dx}&=1\\ \end{aligned}\end{equation} $$
Using the chain rule: $$ \begin{equation}\begin{aligned} \frac{dy}{dx}&=\frac{dy}{du}\times \frac{du}{dx}\\ &=\cos{u}\times 1\\ &=\cos{u}\\ \frac{dy}{dx}&=\cos(x+1)\\ \end{aligned}\end{equation} $$
Example 2
Differentiate the function $$ \begin{equation}\begin{aligned} y=\cos(x^2+2x)\\ \end{aligned}\end{equation} $$
We can let $u$ be the inner function $x^2+2x$: $$ \begin{equation}\begin{aligned} u&=x^2+2x\\ \therefore y&=\cos{u}\\ \end{aligned}\end{equation} $$
Having $y$ in terms of $u$ we find the derivative: $$ \begin{equation}\begin{aligned} \frac{dy}{du}&=-\sin{u}\\ \end{aligned}\end{equation} $$
Now with $u$ written in terms of $x$, the derivative is $$ \begin{equation}\begin{aligned} \frac{du}{dx}&=2x+2\\ \end{aligned}\end{equation} $$
Thus by the chain rule, the final derivative is $$ \begin{equation}\begin{aligned} \frac{dy}{dx}&=\frac{dy}{du}\times \frac{du}{dx}\\ &=-\sin{u}\times (2x+2)\\ \frac{dy}{dx}&=-(2x+2)\sin(x^2+2x)\\ \end{aligned}\end{equation} $$
Example 3
Find the derivative of the function $y=(5x+4)^7$
Here the inner function is $5x+4$ thus we can let this be $u$: $$ \begin{equation}\begin{aligned} u&=5x+4\\ \therefore y&=u^7\\ \end{aligned}\end{equation} $$
Now $y$ is in terms of $u$ which makes the derivative: $$ \begin{equation}\begin{aligned} \frac{dy}{du}&=7u^6\\ \end{aligned}\end{equation} $$
And with $u$ in terms of $x$: $$ \begin{equation}\begin{aligned} \frac{du}{dx}&=5\\ \end{aligned}\end{equation} $$
Therefore the full derivative is $$ \begin{equation}\begin{aligned} \frac{dy}{dx}&=\frac{dy}{du}\times \frac{du}{dx}\\ &=7u^6\times 5\\ &=35u^6\\ \frac{dy}{dx}&=35(5x+4)^6\\ \end{aligned}\end{equation} $$
Chain rule practice
Find the derivatives of the following compositions of functions:
- $y=(x-1)^{10}$
- $y=\sin(3x+4)$
- $y=(x^3-\frac{1}{x})^5$
- $y=\cos(\sqrt{x}+1)$
Product rule
This rule is used to find the derivative of the product of two functions: $$ \begin{equation}\begin{aligned} \frac{d}{dx}[fg]=fg'+gf'\\ \end{aligned}\end{equation} $$
Notice how whether $f$ or $g$, a function is never paired with its own derivative. Thus you will never see $ff’$ or $gg’$ in product rule.
Example
Given the function $$ \begin{equation}\begin{aligned} y=(x+3)(2x+1)\\ \end{aligned}\end{equation} $$
Because we have the product of two functions, we can find the derivative of the first function: $$ \begin{equation}\begin{aligned} \frac{d}{dx}[x+3]=1\\ \end{aligned}\end{equation} $$
And that for the second function: $$ \begin{equation}\begin{aligned} \frac{d}{dx}[2x+1]=2\\ \end{aligned}\end{equation} $$
Applying the product rule: $$ \begin{equation}\begin{aligned} \frac{dy}{dx}&=(x+3)(2) + (2x+1)(1)\\ &=2x+6+2x+1\\ &=4x+7\\ \end{aligned}\end{equation} $$
Note how $(x+3)$ is placed next to the derivative of the second function, $2$, and $(2x+1)$ is placed next to the derivative of the first function, $1$.
Product rule practice
Find the derivatives of the following functions:
- $y=(x+1)\sin{x}$
- $y=(x^2+3x)(-5x+7)$
- $y=\sin{x}\cos{x}$
- $y=(x^3-3x+4)(9x^2-4x)$
- $f(x)=(11\sqrt{x}+2x)(\frac{1}{x}-1)$
Quotient rule
This rule is used to find the derivative of the quotient of two functions: $$ \begin{equation}\begin{aligned} \frac{d}{dx}[\frac{f}{g}]=\frac{gf'-fg'}{g^2}\\ \end{aligned}\end{equation} $$
Note that we begin with the function at the bottom (in this case, $g$).
Example
Given the quotient of two functions $$ \begin{equation}\begin{aligned} y=\frac{x-3}{5x+1}\\ \end{aligned}\end{equation} $$
We find the derivatives $$ \begin{equation}\begin{aligned} \frac{d}{dx}[x-3]&=1\\ \frac{d}{dx}[5x+1]&=5\\ \end{aligned}\end{equation} $$
Applying the quotient rule $$ \begin{equation}\begin{aligned} \frac{dy}{dx}&=\frac{(5x+1)(1)+(x-3)(5)}{(5x+1)^2}\\ &=\frac{5x+1+5x-15}{(5x+1)^2}\\ &=\frac{10x-14}{(5x+1)^2}\\ \end{aligned}\end{equation} $$
Quotient rule practice
Determine the derivatives of the following functions:
- $y=\frac{2x-4}{x+5}$
- $y=\frac{\sin{x}}{x^2-1}$
- $y=\frac{x^4+5x}{7x^3-14}$
- $y=\frac{5\cos{x}}{\sqrt{x}-2}$
- $y=\frac{x}{x^4-8x+15}$