Further Differentiation of Brackets

Video lesson · from 1:51:474 worked examples

One lesson video covers all of Differentiation, so it opens at 1:51:47 for this topic — not from the beginning.

When differentiating composite functions involving linear expressions raised to a power, we can use the chain rule.

If the functions $f$ and $g$ are defined on suitable domains, then:

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

For brackets raised to a power:

\frac{d}{dx}[(ax+b)^n] = n(ax+b)^{n-1} \times a = an(ax+b)^{n-1}

Alternatively, using Leibniz notation, if $y = f(u)$ and $u = g(x)$ :

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Multiply by the derivative of the inside: the chain rule is easy to state and easy to forget. $(3x+1)^5$ differentiates to $5(3x+1)^4 \times 3$ , and that $\times 3$ is where the marks go.
Do not expand the bracket: for a high power the chain rule is the only realistic route.
Keep the inside unchanged: only the power drops. The contents of the bracket stay exactly as they were.
Negative and fractional powers still follow the rule: $\frac{1}{(2x-5)^3}$ should be written $(2x-5)^{-3}$ before differentiating.
Simplify at the end: collect the constants into a single coefficient.

Example 1

Differentiate $y = (5x + 2)^4$ with respect to $x$ .

Using the chain rule:

Let $u = 5x + 2$ , then $\frac{du}{dx} = 5$ .

And $y = u^4$ , then $\frac{dy}{du} = 4u^3$ .

\begin{aligned} \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx} \\ &= 4u^3 \times 5 \\ &= 20(5x + 2)^3 \end{aligned}

Example 2

Differentiate $y = \frac{1}{(5x + 2)^3}$ with respect to $x$ .

First, rewrite the expression with a negative exponent:

y = (5x + 2)^{-3}

Now apply the chain rule:

Let $u = 5x + 2$ , so $\frac{du}{dx} = 5$ .

\begin{aligned} \frac{dy}{dx} &= -3(5x + 2)^{-4} \times 5 \\ &= -15(5x + 2)^{-4} \\ &= -\frac{15}{(5x + 2)^4} \end{aligned}

Example 3

A function $f$ is defined by $f(x) = \frac{5}{(5x + 2)^6}$ for $x \in \mathbb{R}$ . Find $f'(x)$ .

Rewrite to prepare for differentiation:

f(x) = 5(5x + 2)^{-6}

\begin{aligned} f'(x) &= 5 \times -6(5x + 2)^{-7} \times 5 \\ &= -150(5x + 2)^{-7} \\ &= -\frac{150}{(5x + 2)^7} \end{aligned}

Example 4

A function $f$ is defined on a suitable domain by $f(x) = \sqrt{5x^2 + 2x}$ . Find $f'(x)$ .

First, rewrite as a fractional exponent:

f(x) = (5x^2 + 2x)^{1/2}

Now apply the chain rule:

\begin{aligned} f'(x) &= \frac{1}{2}(5x^2 + 2x)^{-1/2} \times (10x + 2) \\ &= \frac{10x + 2}{2\sqrt{5x^2 + 2x}} \\ &= \frac{5x + 1}{\sqrt{5x^2 + 2x}} \end{aligned}