Differentiation

New Zealand

Level 7 - NCEA Level 2

Lesson

Recall that the power rule is

Power Rule

If $f(x)=ax^n$`f`(`x`)=`a``x``n` then $f'(x)=nax^{n-1}$`f`′(`x`)=`n``a``x``n`−1

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:

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`)=−2`x`−2−1=−2`x`−3

If we have terms like $\frac{4}{x}$4`x` or $\frac{-1}{2x}$−12`x`, 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 $\frac{a}{b}$`a``b` we can rewrite it as $ab^{-1}$`a``b`−1.

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

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

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`=`x`12 or that more generally $\sqrt[n]{x}=x^{\frac{1}{n}}$^{n}√`x`=`x`1`n`

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

$f(x)=3\times x^{\frac{1}{3}}$`f`(`x`)=3×`x`13 which means that the derivative is

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

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

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

$f\left(x\right)=\frac{x^2}{2x}-\frac{4x}{2x}+\frac{8}{2x}$`f`(`x`)=`x`22`x`−4`x`2`x`+82`x`

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

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

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

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

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

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

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

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

Find $\frac{dy}{dx}$

`d``y``d``x`.

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

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

`x`.Find the derivative of the function.

Apply differentiation and anti-differentiation techniques to polynomials

Apply calculus methods in solving problems