Differentiating Powers of x (AQA GCSE Further Maths): Revision Note

• Don't try to do too many steps in your head; write the expression in a format that you can differentiate before you actually differentiate it

  • e.g. $y=\frac{1}{x^4}+\frac{2}{x^3}$ can be rewritten as $y=x^{-4}+2x^{-3}$ which is then far easier to differentiate

Find the derivative of 

(a)$\begin{array}{c}\\ \\ \end{array}$ $y=5x^3+2x+\frac{3}{x^2}+8$

Rewrite the $\frac{3}{x^2}$ term

$y=5x^3+2x+3x^{-2}+8$

Apply the rule for differentiating powers ($y=a{x}^n,\frac{\mathrm{d}y}{\mathrm{d}x}=an{x}^{n-1}$) and apply the special cases for the terms $2x$ and 8 ($y=ax,\frac{\mathrm{d}y}{\mathrm{d}x}=a$ and $y=a,\frac{\mathrm{d}y}{\mathrm{d}x}=0$)

$\frac{\mathrm{d}y}{\mathrm{d}x}=15x^2+2-6x^{-3}$

Unless a question specifies there is not usually a need to rewrite/simplify the answer

$\frac{\mathbf{d}\mathit{y}}{\mathbf{d}\mathit{x}}\mathbf{=}\mathbf{15}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{-}\frac{\mathbf{6}}{{\mathit{x}}^{\mathbf{3}}}$

 (b) $y={\left(2x+3\right)}^2$

This is a product of two (equal) brackets so cannot be differentiated directly
Expand the brackets to get an expression in powers of $x$
Take time to get the expansion correct, writing stages out in full if necessary

$\begin{array}{rcl}y& =& \left(2x+3\right)\left(2x+3\right)\\ y& =& 4x^2+6x+6x+9\\ y& =& 4x^2+12x+9\end{array}$

Differentiate 'term-by-term', looking out for those special cases

$\frac{\mathrm{d}y}{\mathrm{d}x}=8x+12$

There is a factor of 4 but there is no demand to factoise the final answer in the question

$\frac{\mathbf{d}\mathit{y}}{\mathbf{d}\mathit{x}}\mathbf{=}\mathbf{8}\mathit{x}\mathbf{+}\mathbf{12}$

(c)$\begin{array}{c}\\ \\ \end{array}$$y=\frac{8x^6-x^3}{2x^4}$

This is a quotient so cannot be differentiated directly
Spot the single denominator which means we can split the fraction by the two terms on the numerator

$y=\frac{8x^6}{2x^4}-\frac{x^3}{2x^4}$

Simplify using the laws of indices

$$\begin{gathered} y=4x^{6-4}-\frac{1}{2}{x}^{3-4} \\ y=4x^2-\frac{1}{2}{x}^{-1} \end{gathered}$$

Each term is now a power of $x$, so differentiate 'term-by-term'

$\frac{\mathrm{d}y}{\mathrm{d}x}=8x+\frac{1}{2}{x}^{-2}$

There is demand to simplify or write the answer in a particular form

$\frac{\mathbf{d}\mathit{y}}{\mathbf{d}\mathit{x}}\mathbf{=}\mathbf{8}\mathit{x}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{x}}^{\mathbf{-}\mathbf{2}}$

• Read the question carefully; sometimes you are given $\frac{dy}{dx}$ and so don't need to differentiate initially - don't just automatically differentiate the first thing you see!

• The following mean the same thing:

  • "Find the gradient of the curve at $x=2$"

  • "Find the gradient of the tangent at $x=2$"

    • the tangent gradient = curve gradient at that point

  • "Find the rate of change of y with respect to x at $x=2$"

A curve has the equation $y=\frac{4}{3}{x}^3+3x-8$.

(a) Find the gradient of the curve when $x=2$.

$y$ is already in a form that can be differentiated

$\frac{\mathrm{d}y}{\mathrm{d}x}=4x^2+3$

Substitute $x=2$ into $\frac{\mathrm{d}y}{\mathrm{d}x}$

$\frac{\mathrm{d}y}{\mathrm{d}x}=4\times 2^2+3=19$

The gradient of the curve at $\mathit{x}\mathbf{=}\mathbf{2}$ is 19

(b)

Work out the possible values of $x$ for which the rate of change of $y$ with respect to $x$ is 4.

"Rate of change" is another way of describing the derivate

$\begin{array}{rcl}\frac{\mathrm{d}y}{\mathrm{d}x}& =& 4\\ 4x^2+3& =& 4\end{array}$

Solve this equation to find $x$
Note that it is quadratic equation so it could have up to two solutions
The question refers to 'values' implying there is (or could be) more than one value for $x$

$\begin{array}{rcl}4x^2& =& 1\\ x^2& =& \frac{1}{4}\\ x& =& \pm \sqrt{\frac{1}{4}}\\ x& =& \pm \frac{1}{2}\end{array}$

The possible values of $\mathit{x}$, that give a rate of change of 4, are $\mathit{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$ and $\mathit{x}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}$