Further derivatives using power rule

Recall that the power rule is

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

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)=x−2. then, by the power rule, the derivative is $f'(x)=-2x^{-2-1}=-2x^{-3}$f′(x)=−2x−2−1=−2x−3

 

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 $ab^{-1}$ab−1.

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

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

 

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 $\sqrt[n]{x}=x^{\frac{1}{n}}$ⁿ√x=x1n​

So, to calculate the derivative of the function $f(x)=3\sqrt[3]{x}$f(x)=3³√x 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

$f'\left(x\right)=\frac{1}{3}\times3x^{\frac{1}{3}-1}=x^{-\frac{2}{3}}$f′(x)=13​×3x13​−1=x−23​

 

Divide and simplify first if necessary

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

We are not so familiar with simplifying expressions like $f\left(x\right)=\frac{x^2-4x+8}{2x}$f(x)=x2−4x+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)=x2−4x+82x​, becomes 

$f\left(x\right)=\frac{x^2}{2x}-\frac{4x}{2x}+\frac{8}{2x}$f(x)=x22x​−4x2x​+82x​

$f\left(x\right)=\frac{x}{2}-2+\frac{4}{x}$f(x)=x2​−2+4x​

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}$4x−1, 

$f\left(x\right)=\frac{x}{2}-2+4x^{-1}$f(x)=x2​−2+4x−1

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

$f'\left(x\right)=\frac{1}{2}-4x^{-2}$f′(x)=12​−4x−2

Practice questions:

question 1

question 2

question 3