# Implicit Differentiation

A LevelAQAEdexcelOCRAQA 2022OCR 2022

## Implicit Differentiation

An implicit relation between $x$ and $y$ is one written as $f(x,y)=g(x,y)$. They often appear for relations that it is impossible to write in the form $y=f(x)$. Despite not having a nice expression for $y$ in terms of $x$, we can still differentiate implicit relations.

A Level

## Method for Implicit Differentiation

To carry out implicit differentiation, follow these steps.

Step 1: Differentiate terms that are in $x$ only.

Step 2: Use the chain rule to differentiate terms in $y$ only.

$\dfrac{d}{dx}(f(y))=\dfrac{d}{dy}(f(y))\dfrac{dy}{dx}$

This is the same as differentiating $f(y)$ normally then multiplying by $\dfrac{dy}{dx}$.

Step 3: Use the product rule for terms that are in both $x$ and $y$.

$\dfrac{d}{dx}(u(x)v(y))=u(x)\dfrac{d}{dx}(v(y))+v(y)\dfrac{d}{dx}(u(x))$

And you can use the rule in step 2 to find $\dfrac{d}{dx}(v(y))$.

Step 4: Rearrange the resulting equation to make $\dfrac{dy}{dx}$ the subject.

A Level

## Finding the Gradient with Implicit Differentiation

Implicit differentiation still finds the gradient. We can calculate the gradient at a given $x$ value by finding the $y$ value from the equation for the curve then finding $\dfrac{dy}{dx}$ at these values of $x$ and $y$.

We can also find other things, such as an equation linking $x$ and $y$ at stationary points: $\dfrac{dy}{dx}=0$.

Example: A curve is defined by $y^{2}-3xy+x^{2}=0$. Find the gradient at all points where $x=2$.

$y^{2}-3xy+x^{2}=0$

Implicitly differentiate:

$2y\dfrac{dy}{dx}-3x\dfrac{dy}{dx}-3y+2x=0$

Rearrange:

$(2y-3x)\dfrac{dy}{dx}-3y+2x=0$

$\dfrac{dy}{dx}+\dfrac{2x-3y}{2y-3x}=0$

$\dfrac{dy}{dx}=\dfrac{3y-2x}{2y-3x}$

Substitute $x$ value into curve to find $y$ values:

At $x=2$, $y^{2}-6y+4=0$

$y=\dfrac{-(-6)\pm\sqrt{(-6)^{2}-4\times1\times4}}{2\times1}$

$y=\dfrac{6\pm\sqrt{36-16}}{2}$

$y=3\pm\dfrac{\sqrt{20}}{2}$

$y=3\pm\sqrt{5}$

Substitute $x,y$ values into $\dfrac{dy}{dx}$:

 $y=3+\sqrt{5}$ $\dfrac{dy}{dx}=\dfrac{3(3+\sqrt{5})-2\times2}{2(3+\sqrt{5})-3\times2}$ $\dfrac{dy}{dx}=\dfrac{9+3\sqrt{5}-4}{6+2\sqrt{5}-6}$ $\dfrac{dy}{dx}=\dfrac{5+3\sqrt{5}}{2\sqrt{5}}$ $\dfrac{dy}{dx}=\dfrac{5\sqrt{5}+15}{10}$ $\dfrac{dy}{dx}=\dfrac{3+\sqrt{5}}{2}$ $y=3-\sqrt{5}$ $\dfrac{dy}{dx}=\dfrac{3(3-\sqrt{5})-2\times2}{2(3-\sqrt{5})-3\times2}$ $\dfrac{dy}{dx}=\dfrac{9-3\sqrt{5}-4}{6-2\sqrt{5}-6}$ $\dfrac{dy}{dx}=\dfrac{5-3\sqrt{5}}{-2\sqrt{5}}$ $\dfrac{dy}{dx}=\dfrac{5\sqrt{5}-15}{-10}$ $\dfrac{dy}{dx}=\dfrac{3-\sqrt{5}}{2}$

Hence, gradients at $x=2$ are $\dfrac{3+\sqrt{5}}{2}$ and $\dfrac{3-\sqrt{5}}{2}$

A Level

## Differentiating the Inverse Trigonometric Functions

Implicit differentiation is how we find the derivative of $\arcsin$, $\arccos$ and $arctan$.

Example: Find the derivative of $\arcsin(x)$.

$y=\arcsin(x)$

Take $\sin$ of both sides:

$\sin(y)=x$

Differentiate (implicitly):

$\cos(y)\dfrac{dy}{dx}=1$

$\dfrac{dy}{dx}=\dfrac{1}{\cos(y)}$

$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{\cos^{2}(y)}}$

We know that $\sin^{2}(y)+\cos^{2}(y)=1$, so $\cos^{2}(y)=1-\sin^{2}(y)$

$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-\sin^{2}(y)}}$

Now $x=\sin(y)$ so we can put this back into the equation.

$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^{2}}}$

A Level
A Level

## Example 1: Implicit Differentiation

$4x^{3}+23\sin(x)\cos(y)+36xy^{2}+12y+1=0$. Find $\dfrac{dy}{dx}$.

[3 marks]

$4x^{3}+23\sin(x)\cos(y)+36xy^{2}+12y+1=0$

Differentiate implicitly.

$12x^{2}+23\cos(x)\cos(y)-23\sin(x)\sin(y)\dfrac{dy}{dx}+36y^{2}+72xy\dfrac{dy}{dx}+12\dfrac{dy}{dx}=0$

$12x^{2}+23\cos(x)\cos(y)+36y^{2}=23\sin(x)\sin(y)\dfrac{dy}{dx}-72xy\dfrac{dy}{dx}-12\dfrac{dy}{dx}$

$12x^{2}+23\cos(x)\cos(y)+36y^{2}=(23\sin(x)\sin(y)-72xy-12)\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{12x^{2}+23\cos(x)\cos(y)+36y^{2}}{23\sin(x)\sin(y)-72xy-12}$

A Level

## Example 2: Inverse Trigonometric Functions

Find the derivative of $\arccos(x)$.

[4 marks]

$y=\arccos(x)$

Take $\cos$ of both sides.

$\cos(y)=x$

Implicitly differentiate.

$-\sin(y)\dfrac{dy}{dx}=1$

$\dfrac{dy}{dx}=-\dfrac{1}{\sin(y)}$

$\dfrac{dy}{dx}=-\dfrac{1}{\sqrt{\sin^{2}(y)}}$

We know $\sin^{2}(y)+\cos^{2}(y)=1$, so $\sin^{2}(y)=1-\cos^{2}(y)$

$\dfrac{dy}{dx}=-\dfrac{1}{\sqrt{1-\cos^{2}(y)}}$

Now $x=\cos(y)$ so we can put this back into the equation.

$\dfrac{dy}{dx}=-\dfrac{1}{\sqrt{1-x^{2}}}$

A Level

## Example Questions

$9\ln(x)\cos(y)=0$

Differentiate implicitly:

$\dfrac{9\cos(y)}{x}-9\ln(x)\sin(y)\dfrac{dy}{dx}=0$

$\dfrac{9\cos(y)}{x}=9\ln(x)\sin(y)\dfrac{dy}{dx}=0$

$\dfrac{dy}{dx}=\dfrac{9\cos(y)}{9x\ln(x)\sin(y)}$

$\dfrac{dy}{dx}=\dfrac{\cos(y)}{x\ln(x)\sin(y)}$
$3x^{2}y-x=0$

Differentiate implicitly.

$6xy+3x^{2}\dfrac{dy}{dx}-1=0$

$3x^{2}\dfrac{dy}{dx}=1-6xy$

$\dfrac{dy}{dx}=\dfrac{1-6xy}{3x^{2}}$

Find $y$ when $x=\dfrac{1}{3}$.

$3\left(\dfrac{1}{3}\right)^{2}y-\dfrac{1}{3}=0$

$3\times\dfrac{1}{9}y-\dfrac{1}{3}=0$

$\dfrac{1}{3}y-\dfrac{1}{3}=0$

$\dfrac{1}{3}y=\dfrac{1}{3}$

$y=1$

Substitute into our expression for $\dfrac{dy}{dx}$:

$\dfrac{dy}{dx}=\dfrac{1-6\times1\times\dfrac{1}{3}}{3\left(\dfrac{1}{3}\right)^{2}}$

$\dfrac{dy}{dx}=\dfrac{1-2}{3\times\dfrac{1}{9}}$

$\dfrac{dy}{dx}=\dfrac{-1}{\dfrac{1}{3}}$

$\dfrac{dy}{dx}=-3$

So the gradient is $-3$ at this point.

a) $3x^{3}y^{2}-xy=1$

$9x^{2}y^{2}+6x^{3}y\dfrac{dy}{dx}-y-x\dfrac{dy}{dx}=0$

$9x^{2}y^{2}-y=x\dfrac{dy}{dx}-6x^{3}y\dfrac{dy}{dx}$

$9x^{2}y^{2}-y=(x-6x^{3}y)\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{9x^{2}y^{2}-y}{x-6x^{3}y}$

b) Stationary point is at $\dfrac{dy}{dx}=0$

$0=\dfrac{9x^{2}y^{2}-y}{x-6x^{3}y}$

$9x^{2}y^{2}-y=0$

$y(9x^{2}y-1)=0$

$y=0$ or $9x^{2}y-1=0$

$y=0$ or $9x^{2}y=1$

c) Curve equation is $3x^{3}y^{2}-xy=1$

$y=0$ gives $0=1$ so we cannot use this result.

$9x^{2}y=1$

$y=\dfrac{1}{9x^{2}}$

Substitute this into the curve.

$3x^{3}\left(\dfrac{1}{9x^{2}}\right)^{2}-\dfrac{x}{9x^{2}}=1$

$\dfrac{3x^{3}}{81x^{4}}-\dfrac{1}{9x}=1$

$\dfrac{1}{27x}-\dfrac{1}{9x}=1$

$1-3=27x$

$x=-\dfrac{2}{27}$

$y=\dfrac{1}{9x^{2}}$

$y=\dfrac{1}{9\left(-\dfrac{2}{27}\right)^{2}}$

$y=\dfrac{1}{9\left(\dfrac{4}{729}\right)}$

$y=\dfrac{1}{\left( \dfrac{4}{81}\right) }$

$y=\dfrac{81}{4}$

So the stationary point is $\left(-\dfrac{2}{27},\dfrac{81}{4}\right)$

$y=\arctan(x)$

Take $\tan$ of both sides.

$\tan(y)=x$

Differentiate implicitly:

$\sec^{2}(y)\dfrac{dy}{dx}=1$

$\dfrac{dy}{dx}=\dfrac{1}{\sec^{2}(y)}$

Identity: $\sec^{2}(y)=1+\tan^{2}(y)$

$\dfrac{dy}{dx}=\dfrac{1}{1+\tan^{2}(y)}$

Put $x=\tan(y)$ back into the equation.

$\dfrac{dy}{dx}=\dfrac{1}{1+x^{2}}$

A Level

A Level

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