Www.le.ac.uk Implicit Differentiation Department of Mathematics University of Leicester.

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Implicit Differentiation

Department of MathematicsUniversity of Leicester

What is it?

• The normal function we work with is a function of the form:

• This is called an Explicit function.

Example:

)(xfy

32 xy

What is it?

• An Implicit function is a function of the form:

• Example:

0),( yxf

0343 xyxy

What is it?

• This also includes any function that can be rearranged to this form: Example:

• Can be rearranged to:

• Which is an implicit function.

yxyxy 2343

02343 yxyxy

What is it?

• These functions are used for functions that would be very complicated to rearrange to the form y=f(x).

• And some that would be possible to rearrange are far to complicated to differentiate.

Implicit differentiation

• To differentiate a function of y with respect to x, we use the chain rule.

• If we have: , and

• Then using the chain rule we can see that:

)(yzz

dx

dy

dy

dz

dx

dz

)(xyy

Implicit differentiation

• From this we can take the fact that for a function of y:

dx

dyyf

dy

dxf

dx

d ))(())((

Implicit differentiation: example 1

• We can now use this to solve implicit differentials

• Example:

yxyxy 2343


• Then differentiate all the terms with respect to x:

)2()()()()( 343 ydx

dx

dx

dy

dx

dx

dx

dy

dx

d


• Terms that are functions of x are easy to differentiate, but functions of y, you need to use the chain rule.


• Therefore:

• And:

dx

dyy

dx

dyy

dy

dy

dx

d 233 3)()(

dx

dyy

dx

dyy

dy

dy

dx

d 344 4)()(

dx

dy

dx

dyy

dy

dy

dx

d2)2()2(


• Then we can apply this to the original function:

• Equals:

)2()()()()( 343 ydx

dx

dx

dy

dx

dx

dx

dy

dx

d

dx

dyx

dx

dyy

dx

dyy 23413 232


• Next, we can rearrange to collect the terms:

• Equals:

dx

dy

dx

dyx

dx

dyy

dx

dyy 23413 232

dx

dyyyx )234(13 232


• Finally:

• Equals:

234

1323

2

yy

x

dx

dy

dx

dyyyx )234(13 232


• Example:

• Differentiate with respect to x

0sin4cos ydx

dxy

dx

d

0sin2cos 2 yxy


• Then we can see, using the chain rule:

• And:

dx

dyy

dx

dyy

dy

dy

dx

dsin)(coscos

dx

dyy

dx

dyy

dy

dy

dx

dcos)(sinsin


• We can then apply these to the function:

• Equals:

0sin4cos ydx

dxy

dx

d

0cos4sin dx

dyyx

dx

dyy


• Now collect the terms:

• Which then equals:

dx

dyyyx )cos(sin4

dx

dy

yy

x

dx

dy

cossin

4


• Example:


22322 yyxx

22322

dx

dy

dx

dyx

dx

dx

dx

d


• To solve:

• We also need to use the product rule.

)()( 233232 x

dx

dyy

dx

dxyx

dx

d

32 yxdx

d

xydx

dyyx 2.3. 322



• Equals:

22322

dx

dy

dx

dyx

dx

dx

dx

d

022.3.2 322 dx

dyyxy

dx

dyyxx




dx

dy

dx

dyyxyxxy )32()22( 223

22

3

32

22

yxy

xxy

dx

dy


• Example:


xyyx 2)arcsin( 22

xdx

dy

dx

dx

dx

d2)arcsin( 22


• To solve:

• We also need to use the chain rule, and the fact that:

)arcsin( 2ydx

d

21

1)arcsin(

xx

dx

d


• This is because:

, so

• Therefore:

• As x=sin(y), Therefore:

21

1)arcsin(

xx

dx

d

dx

dy

)arcsin(xy )sin(yx

22 1sin1cos xyydy

dx


• Using this, we can apply it to our equation :

)arcsin( 2ydx

d

dx

dy

yy

dx

dy

yy

422 1

1.2

)(1

1.2



• Equals:

xdx

dy

dx

dx

dx

d2)arcsin( 22

21

22

4

dx

dy

y

yx




dx

dy

dx

dy

y

yx

41

222

y

yx

dx

dy

2

)1)(22( 4

Conclusion

• Explicit differentials are of the form:

• Implicit differentials are of the form:

)(xfy

0),( yxf

Conclusion

• To differentiate a function of y with respect to x we can use the chain rule:

dx

dyyf

dy

dxf

dx

d ))(())((

Www.le.ac.uk Implicit Differentiation Department of Mathematics University of Leicester.

Documents

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