Implicit Differentiation

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 AQA Edexcel OCR

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 LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

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 LevelAQAEdexcelOCR

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 LevelAQAEdexcelOCR

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}}

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