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

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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

Question 1: A curve is $9\ln(x)\cos(y)=0$ . Find $\dfrac{dy}{dx}$ .

[2 marks]

A Level

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)}

Question 2: Find the gradient of the curve $3x^{2}y-x=0$ when $x=\dfrac{1}{3}$

[3 marks]

A Level

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.

Question 3:

a) Find the implicit derivative of $3x^{3}y^{2}-xy=1$

b) Find an equation linking $x$ and $y$ at the stationary points of the curve.

c) Use this equation and the equation of the curve to find the stationary points of the curve.

[8 marks]

A Level

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 $\left(-\dfrac{2}{27},\dfrac{81}{4}\right)$

Question 4: Find the derivative of $y=\arctan(x)$ .

[3 marks]

A Level

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}}

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