topic badge
New Zealand
Level 7 - NCEA Level 2

Further derivatives using power rule


Recall that the power rule is

Power Rule

If $f(x)=ax^n$f(x)=axn then $f'(x)=nax^{n-1}$f(x)=naxn1

also, recall that the derivative of sum is equal to the sum of the derivatives. 

Sum of Derivatives

Derivative of sum is equal to the sum of the derivatives.

If $f(x)=g(x)\pm h(x)$f(x)=g(x)±h(x) then $f'(x)=g'(x)\pm h'(x)$f(x)=g(x)±h(x)

This means we can apply the power rule to individual terms. 

The key ideas we need to remember when working with the power rule are:

Watch for negatives in the exponents

Now I have mentioned this before, but it is so important I'm going to mention it again. It is such a common mistake that students make and I want to make sure you don't do it! 

Be careful finding the derivative for terms with negative indices. 

For example, if $f(x)=x^{-2}$f(x)=x2. then, by the power rule, the derivative is $f'(x)=-2x^{-2-1}=-2x^{-3}$f(x)=2x21=2x3


Turn fractions into powers

If we have terms like $\frac{4}{x}$4x or $\frac{-1}{2x}$12x, to use the power rule these will first need to be converted to powers.

Recall from our work on negative indices, that for a fraction $\frac{a}{b}$ab we can rewrite it as $ab^{-1}$ab1.

So $\frac{4}{x}$4x would become $4x^{-1}$4x1 and its derivate is therefore $-4x^{-2}$4x2

and $\frac{-1}{2x}$12x becomes $\frac{-1}{2}x^{-1}$12x1 with a derivative of $\frac{1}{2}x^{-2}$12x2


Turn all other $x$x terms into powers

Sometimes there is work to be done with other terms that have the required variable in them.  Terms like square roots or cube roots for example. 

Recall that $\sqrt{x}=x^{\frac{1}{2}}$x=x12  or that more generally $\sqrt[n]{x}=x^{\frac{1}{n}}$nx=x1n

So, to calculate the derivative of the function $f(x)=3\sqrt[3]{x}$f(x)=33x we need to first turn it into a power term.

$f(x)=3\times x^{\frac{1}{3}}$f(x)=3×x13 which means that the derivative is



Divide and simplify first if necessary

We are all very familiar with expanding expressions like $2x(3x-7)$2x(3x7) . We know that every term in the bracket is multiplied by the term outside the brackets i.e $2x(3x-7)=6x2-14x$2x(3x7)=6x214x

We are not so familiar with simplifying expressions like $f\left(x\right)=\frac{x^2-4x+8}{2x}$f(x)=x24x+82x .  The same rules apply as the fraction bar (vinculum) acts like brackets, so we need to divide every term in the numerator by the denominator.

So $f\left(x\right)=\frac{x^2-4x+8}{2x}$f(x)=x24x+82x, becomes 



from here, if we wished to find the derivative we would need to change the $\frac{4}{x}$4x to a power term, which is $4x^{-1}$4x1


The derivative $f'(x)$f(x) is therefore, 


Practice questions:

question 1

Differentiate $y=\frac{2}{\sqrt{x}}$y=2x, giving your final answer in surd form.

question 2

Consider the function $y=\frac{5x\sqrt{x}}{4x^5}$y=5xx4x5.

  1. Fully simplify the function, expressing your answer with a negative index.

  2. Find $\frac{dy}{dx}$dydx.

question 3

Consider the function $y=\frac{8x^9-4x^8+6x^7+9}{2x^2}$y=8x94x8+6x7+92x2.

  1. Rewrite the function so that each term is a power of $x$x.

  2. Find the derivative of the function.




Apply differentiation and anti-differentiation techniques to polynomials


Apply calculus methods in solving problems

What is Mathspace

About Mathspace