Implicit Differentiation

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Implicit Differentiation

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

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Method for Implicit Differentiation

To carry out implicit differentiation, follow these steps.

Step 1: Differentiate terms that are in xx only.

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

ddx(f(y))=ddy(f(y))dydxdfrac{d}{dx}(f(y))=dfrac{d}{dy}(f(y))dfrac{dy}{dx}

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

Step 3: Use the product rule for terms that are in both xx and yy.

ddx(u(x)v(y))=u(x)ddx(v(y))+v(y)ddx(u(x))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 ddx(v(y))dfrac{d}{dx}(v(y)).

Step 4: Rearrange the resulting equation to make dydxdfrac{dy}{dx} the subject.

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Finding the Gradient with Implicit Differentiation

Implicit differentiation still finds the gradient. We can calculate the gradient at a given xx value by finding the yy value from the equation for the curve then finding dydxdfrac{dy}{dx} at these values of xx and yy.

We can also find other things, such as an equation linking xx and yy at stationary points: dydx=0dfrac{dy}{dx}=0.

Example: A curve is defined by y2−3xy+x2=0y^{2}-3xy+x^{2}=0. Find the gradient at all points where x=2x=2.

y2−3xy+x2=0y^{2}-3xy+x^{2}=0

Implicitly differentiate:

2ydydx−3xdydx−3y+2x=02ydfrac{dy}{dx}-3xdfrac{dy}{dx}-3y+2x=0

Rearrange:

(2y−3x)dydx−3y+2x=0(2y-3x)dfrac{dy}{dx}-3y+2x=0

dydx+2x−3y2y−3x=0dfrac{dy}{dx}+dfrac{2x-3y}{2y-3x}=0

dydx=3y−2x2y−3xdfrac{dy}{dx}=dfrac{3y-2x}{2y-3x}

Substitute xx value into curve to find yy values:

At x=2x=2, y2−6y+4=0y^{2}-6y+4=0

y=−(−6)±(−6)2−4×1×42×1y=dfrac{-(-6)pmsqrt{(-6)^{2}-4times1times4}}{2times1}

y=6±36−162y=dfrac{6pmsqrt{36-16}}{2}

y=3±202y=3pmdfrac{sqrt{20}}{2}

y=3±5y=3pmsqrt{5}

Substitute x,yx,y values into dydxdfrac{dy}{dx}:

y=3+5y=3+sqrt{5}

dydx=3(3+5)−2×22(3+5)−3×2dfrac{dy}{dx}=dfrac{3(3+sqrt{5})-2times2}{2(3+sqrt{5})-3times2}

dydx=9+35−46+25−6dfrac{dy}{dx}=dfrac{9+3sqrt{5}-4}{6+2sqrt{5}-6}

dydx=5+3525dfrac{dy}{dx}=dfrac{5+3sqrt{5}}{2sqrt{5}}

dydx=55+1510dfrac{dy}{dx}=dfrac{5sqrt{5}+15}{10}

dydx=3+52dfrac{dy}{dx}=dfrac{3+sqrt{5}}{2}

y=3−5y=3-sqrt{5}

dydx=3(3−5)−2×22(3−5)−3×2dfrac{dy}{dx}=dfrac{3(3-sqrt{5})-2times2}{2(3-sqrt{5})-3times2}

dydx=9−35−46−25−6dfrac{dy}{dx}=dfrac{9-3sqrt{5}-4}{6-2sqrt{5}-6}

dydx=5−35−25dfrac{dy}{dx}=dfrac{5-3sqrt{5}}{-2sqrt{5}}

dydx=55−15−10dfrac{dy}{dx}=dfrac{5sqrt{5}-15}{-10}

dydx=3−52dfrac{dy}{dx}=dfrac{3-sqrt{5}}{2}

Hence, gradients at x=2x=2 are 3+52dfrac{3+sqrt{5}}{2} and 3−52dfrac{3-sqrt{5}}{2}

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Differentiating the Inverse Trigonometric Functions

Implicit differentiation is how we find the derivative of arcsin⁡arcsin, arccos⁡arccos and arctanarctan.

Example: Find the derivative of arcsin⁡(x)arcsin(x).

y=arcsin⁡(x)y=arcsin(x)

Take sin⁡sin of both sides:

sin⁡(y)=xsin(y)=x

Differentiate (implicitly):

cos⁡(y)dydx=1cos(y)dfrac{dy}{dx}=1

dydx=1cos⁡(y)dfrac{dy}{dx}=dfrac{1}{cos(y)}

dydx=1cos⁡2(y)dfrac{dy}{dx}=dfrac{1}{sqrt{cos^{2}(y)}}

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

dydx=11−sin⁡2(y)dfrac{dy}{dx}=dfrac{1}{sqrt{1-sin^{2}(y)}}

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

dydx=11−x2dfrac{dy}{dx}=dfrac{1}{sqrt{1-x^{2}}}

 

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Example 1: Implicit Differentiation

4x3+23sin⁡(x)cos⁡(y)+36xy2+12y+1=04x^{3}+23sin(x)cos(y)+36xy^{2}+12y+1=0. Find dydxdfrac{dy}{dx}.

[3 marks]

4x3+23sin⁡(x)cos⁡(y)+36xy2+12y+1=04x^{3}+23sin(x)cos(y)+36xy^{2}+12y+1=0

Differentiate implicitly.

12x2+23cos⁡(x)cos⁡(y)−23sin⁡(x)sin⁡(y)dydx+36y2+72xydydx+12dydx=012x^{2}+23cos(x)cos(y)-23sin(x)sin(y)dfrac{dy}{dx}+36y^{2}+72xydfrac{dy}{dx}+12dfrac{dy}{dx}=0

12x2+23cos⁡(x)cos⁡(y)+36y2=23sin⁡(x)sin⁡(y)dydx−72xydydx−12dydx12x^{2}+23cos(x)cos(y)+36y^{2}=23sin(x)sin(y)dfrac{dy}{dx}-72xydfrac{dy}{dx}-12dfrac{dy}{dx}

12x2+23cos⁡(x)cos⁡(y)+36y2=(23sin⁡(x)sin⁡(y)−72xy−12)dydx12x^{2}+23cos(x)cos(y)+36y^{2}=(23sin(x)sin(y)-72xy-12)dfrac{dy}{dx}

dydx=12x2+23cos⁡(x)cos⁡(y)+36y223sin⁡(x)sin⁡(y)−72xy−12dfrac{dy}{dx}=dfrac{12x^{2}+23cos(x)cos(y)+36y^{2}}{23sin(x)sin(y)-72xy-12}

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Example 2: Inverse Trigonometric Functions

Find the derivative of arccos⁡(x)arccos(x).

[4 marks]

y=arccos⁡(x)y=arccos(x)

Take cos⁡cos of both sides.

cos⁡(y)=xcos(y)=x

Implicitly differentiate.

−sin⁡(y)dydx=1-sin(y)dfrac{dy}{dx}=1

dydx=−1sin⁡(y)dfrac{dy}{dx}=-dfrac{1}{sin(y)}

dydx=−1sin⁡2(y)dfrac{dy}{dx}=-dfrac{1}{sqrt{sin^{2}(y)}}

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

dydx=−11−cos⁡2(y)dfrac{dy}{dx}=-dfrac{1}{sqrt{1-cos^{2}(y)}}

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

dydx=−11−x2dfrac{dy}{dx}=-dfrac{1}{sqrt{1-x^{2}}}

 

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Implicit Differentiation Example Questions

Question 1: A curve is 9ln⁡(x)cos⁡(y)=09ln(x)cos(y)=0. Find dydxdfrac{dy}{dx}.

[2 marks]

A Level AQAEdexcelOCR

9ln⁡(x)cos⁡(y)=09ln(x)cos(y)=0

 

Differentiate implicitly:

 

9cos⁡(y)x−9ln⁡(x)sin⁡(y)dydx=0dfrac{9cos(y)}{x}-9ln(x)sin(y)dfrac{dy}{dx}=0

 

9cos⁡(y)x=9ln⁡(x)sin⁡(y)dydx=0dfrac{9cos(y)}{x}=9ln(x)sin(y)dfrac{dy}{dx}=0

 

dydx=9cos⁡(y)9xln⁡(x)sin⁡(y)dfrac{dy}{dx}=dfrac{9cos(y)}{9xln(x)sin(y)}

 

dydx=cos⁡(y)xln⁡(x)sin⁡(y)dfrac{dy}{dx}=dfrac{cos(y)}{xln(x)sin(y)}

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Question 2: Find the gradient of the curve 3x2y−x=03x^{2}y-x=0 when x=13x=dfrac{1}{3}

[3 marks]

A Level AQAEdexcelOCR

3x2y−x=03x^{2}y-x=0

 

Differentiate implicitly.

 

6xy+3x2dydx−1=06xy+3x^{2}dfrac{dy}{dx}-1=0

 

3x2dydx=1−6xy3x^{2}dfrac{dy}{dx}=1-6xy

 

dydx=1−6xy3x2dfrac{dy}{dx}=dfrac{1-6xy}{3x^{2}}

 

Find yy when x=13x=dfrac{1}{3}.

 

3(13)2y−13=03left(dfrac{1}{3}right)^{2}y-dfrac{1}{3}=0

 

3×19y−13=03timesdfrac{1}{9}y-dfrac{1}{3}=0

 

13y−13=0dfrac{1}{3}y-dfrac{1}{3}=0

 

13y=13dfrac{1}{3}y=dfrac{1}{3}

 

y=1y=1

 

Substitute into our expression for dydxdfrac{dy}{dx}:

 

dydx=1−6×1×133(13)2dfrac{dy}{dx}=dfrac{1-6times1timesdfrac{1}{3}}{3left(dfrac{1}{3}right)^{2}}

 

dydx=1−23×19dfrac{dy}{dx}=dfrac{1-2}{3timesdfrac{1}{9}}

 

dydx=−113dfrac{dy}{dx}=dfrac{-1}{dfrac{1}{3}}

 

dydx=−3dfrac{dy}{dx}=-3

 

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

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Question 3:

a) Find the implicit derivative of 3x3y2−xy=13x^{3}y^{2}-xy=1

 

b) Find an equation linking xx and yy 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 AQAEdexcelOCR

a) 3x3y2−xy=13x^{3}y^{2}-xy=1

 

9x2y2+6x3ydydx−y−xdydx=09x^{2}y^{2}+6x^{3}ydfrac{dy}{dx}-y-xdfrac{dy}{dx}=0

 

9x2y2−y=xdydx−6x3ydydx9x^{2}y^{2}-y=xdfrac{dy}{dx}-6x^{3}ydfrac{dy}{dx}

 

9x2y2−y=(x−6x3y)dydx9x^{2}y^{2}-y=(x-6x^{3}y)dfrac{dy}{dx}

 

dydx=9x2y2−yx−6x3ydfrac{dy}{dx}=dfrac{9x^{2}y^{2}-y}{x-6x^{3}y}

 

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

 

0=9x2y2−yx−6x3y0=dfrac{9x^{2}y^{2}-y}{x-6x^{3}y}

 

9x2y2−y=09x^{2}y^{2}-y=0

 

y(9x2y−1)=0y(9x^{2}y-1)=0

 

y=0y=0 or 9x2y−1=09x^{2}y-1=0

 

y=0y=0 or 9x2y=19x^{2}y=1

 

c) Curve equation is 3x3y2−xy=13x^{3}y^{2}-xy=1

 

y=0y=0 gives 0=10=1 so we cannot use this result.

 

9x2y=19x^{2}y=1

 

y=19x2y=dfrac{1}{9x^{2}}

 

Substitute this into the curve.

 

3x3(19x2)2−x9x2=13x^{3}left(dfrac{1}{9x^{2}}right)^{2}-dfrac{x}{9x^{2}}=1

 

3x381x4−19x=1dfrac{3x^{3}}{81x^{4}}-dfrac{1}{9x}=1

 

127x−19x=1dfrac{1}{27x}-dfrac{1}{9x}=1

 

1−3=27x1-3=27x

 

x=−227x=-dfrac{2}{27}

 

y=19x2y=dfrac{1}{9x^{2}}

 

y=19(−227)2y=dfrac{1}{9left(-dfrac{2}{27}right)^{2}}

 

y=19(4729)y=dfrac{1}{9left(dfrac{4}{729}right)}

 

y=1(481)y=dfrac{1}{left( dfrac{4}{81}right) }

 

y=814y=dfrac{81}{4}

 

So the stationary point is (−227,814)left(-dfrac{2}{27},dfrac{81}{4}right)

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

[3 marks]

A Level AQAEdexcelOCR

y=arctan⁡(x)y=arctan(x)

 

Take tan⁡tan of both sides.

 

tan⁡(y)=xtan(y)=x

 

Differentiate implicitly:

 

sec⁡2(y)dydx=1sec^{2}(y)dfrac{dy}{dx}=1

 

dydx=1sec⁡2(y)dfrac{dy}{dx}=dfrac{1}{sec^{2}(y)}

 

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

 

dydx=11+tan⁡2(y)dfrac{dy}{dx}=dfrac{1}{1+tan^{2}(y)}

 

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

 

dydx=11+x2dfrac{dy}{dx}=dfrac{1}{1+x^{2}}

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G5 – Differentiate simple functions and relations defined implicitly or parametrically, for first derivative only

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