Applications of Differentiation (Cambridge (CIE) IGCSE Additional Maths): Revision Note

• 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 derivative

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

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 $\mathit{x}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}$

• In an exam, if you need to show a function is increasing or decreasing you can use either strict (<, >) or non-strict (≤, ≥) inequalities

  • You will get the marks either way in this course

For what values of $x$ is $y=2x^3-3x^2+5$ a decreasing function?

The function is decreasing when its gradient $\left(\frac{dy}{dx}\right)$ is less than 0.

Find the derivative of the function by differentiating.

$\begin{array}{rcl}\frac{dy}{dx}& =& 6x^2-6x\end{array}$ 

Solve the inequality $\frac{dy}{dx}<0$ to find the set of values where the gradient is negative.

$\begin{array}{rcl}6x^2-6x& <& 0\end{array}$

Factorise.

$\begin{array}{rcl}6x\left(x-1\right)& <& 0\end{array}$

The solutions to $\frac{dy}{dx}=0$ are $x=0$ and  $x=1$. Find the correct way around for the inequalities by considering the graph of $\frac{dy}{dx}$. The graph is a positive quadratic, so the function is negative between the values of 0 and 1 (where it is below the $x$-axis).

Considering a sketch of the graph of the gradient function may help you see this.

You can check your answers by considering a sketch of the original function, it should be decreasing at the point where $0<x<1.$

$\mathbf{0}\mathbf{<}\mathit{x}\mathbf{<}\mathbf{1}$

• To be successful in this topic, first make sure you are confident with finding the equation of a straight line!

The function $\mathrm{f}\left(x\right)$ is defined by

 $\mathrm{f}\left(x\right)=2x^4+\frac{3}{x^2} x\ne 0$

a) Find an equation for the tangent to the curve $y=\mathrm{f}\left(x\right)$ at the point where $x=1$, giving your answer in the form $y=mx+c$.

First find the derivative $\mathrm{f}\text{'}\left(x\right)$ by differentiating

Start by rewriting $\mathrm{f}\left(x\right)$as powers of x

$\mathrm{f}\left(x\right)=2x^4+3x^{-2}$

Now differentiate

$\mathrm{f}\text{'}\left(x\right)=8x^3-6x^{-3}$

Now substitute $x=1$ into $\mathrm{f}\text{'}\left(x\right)$ to find the gradient of the tangent

$\begin{array}{rcl}\mathrm{f}\text{'}\left(1\right)& =& 8{\left(1\right)}^3-6{\left(1\right)}^{-3}\\ & =& 8-6\\ & =& 2\end{array}$

We also need the y-coordinate, so substitute $x=1$ into $\mathrm{f}\left(x\right)$ also

$\begin{array}{rcl}\mathrm{f}\left(1\right)& =& 2{\left(1\right)}^4+3{\left(1\right)}^{-2}\\ & =& 2+3\\ & =& 5\end{array}$

Now we can substitute the point (1, 5) and the gradient, 2, into $y-y_1=m\left(x_1\right)\left(x-x_1\right)$

$\begin{array}{rcl}y-5& =& 2(x-1)\end{array}$

Note that we are asked for the final answer in the form $y=mx+c$, so rearrange to this form

$\begin{array}{rcl}y-5& =& 2x-2\end{array}$

$\mathit{y}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{3}$

b) Find an equation for the normal at the point where $x=1$, giving your answer in the form $ax+by+d=0$, where $a$, $b$ and $d$ are integers.

We already have the gradient of the tangent; the gradient of the normal is $\frac{-1}{m}$ where $m$ is the gradient of the tangent

gradient of normal = $\frac{-1}{2}$

Substitute the point (1, 5) from part (a) and the gradient of the normal into $y-y_1=m\left(x-x_1\right)$

$\begin{array}{rcl}y-5& =& \left(\frac{-1}{2}\right)\left(x-1\right)\end{array}$

And rearrange into the form required

$\begin{array}{rcl}y-5& =& -\frac{1}{2}x+\frac{1}{2}\\ \frac{1}{2}x+y-5\frac{1}{2}& =& 0\end{array}$

Note that $a$, $b$ and $d$ must be integers so multiply by 2 to clear the fractions

$\begin{array}{rcl}\mathit{x}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{-}\mathbf{11}& \mathbf{=}& \mathbf{0}\end{array}$