Finding the derivative of a function enables us to determine the slope of the tangent at any point on the function. This provides the instantaneous change in the function at a point. Additionally, this allows us to find equations of tangents and normals at a point and delve a little deeper into the nature and characteristics of different functions.
Remember, when differentiating, the structure of the function is very important and determines which method/s to use. It is up to us to choose the most efficient method.
This lesson will provide an opportunity to review the differentiation rules learnt so far and practise selecting and implementing the right method.
Differentiation of $f(x)=ax^n$f(x)=axn
Used to differentiate individual terms of functions such as $y=x^4$y=x4 or $y=-7x^5$y=−7x5.
For a function $f(x)=ax^n$f(x)=axn, where $n$n can be positive or negative, integer or fraction, the derivative is:
$f'(x)=anx^{n-1}$f′(x)=anxn−1
Chain rule
Used to differentiate powers of a function, for example, $y=\left(3x^2-2\right)^5$y=(3x2−2)5. We could expand this function and use the rule above but it would be onerous and prone to error. Instead, we use the chain rule.
If $y=g(u)$y=g(u) and $u=f(x)$u=f(x) then:
$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$dydx=dydu·dudx
Or, a slightly more specific method for $y=(f(x))^n$y=(f(x))n:
$\frac{dy}{dx}=nf'(x)[f(x)]^{n-1}$dydx=nf′(x)[f(x)]n−1
Product rule
Used to differentiate where there are functions being multiplied together, for example $y=\left(4x-1\right)\left(2x^3-7\right)^4$y=(4x−1)(2x3−7)4. Again, we could expand this but it is much simpler to use the product rule.
If $y=u(x)v(x)$y=u(x)v(x) then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$dydx=udvdx+vdudx
Or:
If $y=uv$y=uv then $y'=uv'+vu'$y′=uv′+vu′
Quotient rule
Used to differentiate when you have one function divided by another, for example $y=\frac{2x-1}{\left(3x^2+2\right)^3}$y=2x−1(3x2+2)3. We could convert the denominator to a term involving negative indices and use the product rule, however, the quotient rule simplifies the process significantly.
If a function is of the form $y=\frac{u}{v}$y=uv, where $u$u and $v$v are functions of $x$x, then
$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$dydx=vdudx−udvdxv2
Or, more simply:
$y'=\frac{vu'-uv'}{v^2}$y′=vu′−uv′v2
Be very careful with subtraction in the numerator - this often creates an expansion with a negative coefficient that needs to be treated carefully to avoid errors.
If there are multiple functions within a function, such as $f(x)=x^3\left(-2x^4+3\right)^4+\frac{7x+3}{(x-6)^4}$f(x)=x3(−2x4+3)4+7x+3(x−6)4, we just apply combinations of the rules above, to the individual terms of the overall function, and add or subtract the resulting derivatives.
Take each question slowly, working step by step, setting out as much work as possible, so that errors (if you make them) are easier to identify.
Practice questions
Question 1
Find the derivative of $y=\left(5x\right)^3+6\sqrt[3]{x}$y=(5x)3+63√x.
Question 2
Differentiate $y=x^3\left(5x+3\right)^7$y=x3(5x+3)7 using the product rule. Express your answer in factorised form.
You may let $u=x^3$u=x3 and $v=\left(5x+3\right)^7$v=(5x+3)7.
Question 3
Consider the function $f\left(t\right)=\frac{\left(4t^2+3\right)^3}{\left(5+2t\right)^5}$f(t)=(4t2+3)3(5+2t)5.
Find $f'\left(t\right)$f′(t).
You may use the substitutions $u=\left(4t^2+3\right)^3$u=(4t2+3)3 and $v=\left(5+2t\right)^5$v=(5+2t)5 in your working.