Perplex
  • Dashboard
Topics
Exponents & LogarithmsRounding & ErrorSequences & SeriesCounting & BinomialsProof and ReasoningComplex NumbersAlgebra Skills
Cartesian plane & linesQuadraticsFunction TheoryTransformations & asymptotesPolynomials
2D & 3D GeometryTrig equations & identitiesVectors
ProbabilityDescriptive StatisticsBivariate StatisticsDistributions & Random Variables
DifferentiationIntegrationDifferential EquationsMaclaurin
Paper 3
Plus
Calculator Skills
Review VideosFormula BookletAll Study Sets
BlogLanding Page
Sign UpLogin
Perplex
Perplex
  • Dashboard
Topics
Exponents & LogarithmsRounding & ErrorSequences & SeriesCounting & BinomialsProof and ReasoningComplex NumbersAlgebra Skills
Cartesian plane & linesQuadraticsFunction TheoryTransformations & asymptotesPolynomials
2D & 3D GeometryTrig equations & identitiesVectors
ProbabilityDescriptive StatisticsBivariate StatisticsDistributions & Random Variables
DifferentiationIntegrationDifferential EquationsMaclaurin
Paper 3
Plus
Calculator Skills
Review VideosFormula BookletAll Study Sets
BlogLanding Page
Sign UpLogin
Perplex

Differentiation (Lesson 6/9)

Implicit differentiation

1 / 6

Consider the equation

​
2y2+x2=1
​

This is not a function, but rather a relation. Its graph looks like this:

But if you drag the point around, you see that the curve still has a well defined slope and tangent! How do we find it?


Since both sides of the equation are equal, they both have to change by the same amount when we make a small change to ​x.

​
dxd​(2y2+x2)=dxd​(1)
​

We know what the derivative of ​x2​ is:

​
(x2)′=2x
​

And ​dxd​(1)=0​ because ​1​ is a constant.


But what about ​2y2? Well since ​y​ depends on ​x, we use the chain rule:

​
dxd​(2y2)=2⋅dyd(y2)​⋅dxdy​
​

But ​dyd(y2)​=2y, so

​
dxd​(2y2)=4y⋅dxdy​
​

Plugging everything back in gives:

​
4y⋅dxdy​+2x=0
​

We can rearrange to find

​
dxdy​=4y2x​=yx​
​


This is an example of a technique called implicit differentiation. It probably seems super weird right now, but after a few practice examples this will start to feel routine.


You might notice that in this case ​dxdy​=yx​, which depends on ​x​ and ​y. This is often the case in implicit differentiation, simply because the curve is not a pure function of ​x, so the slope is usually not either (I say usually because sometimes the curves are symmetric).

Differentiation (Lesson 6/9)

Implicit differentiation

1 / 6

Consider the equation

​
2y2+x2=1
​

This is not a function, but rather a relation. Its graph looks like this:

But if you drag the point around, you see that the curve still has a well defined slope and tangent! How do we find it?


Since both sides of the equation are equal, they both have to change by the same amount when we make a small change to ​x.

​
dxd​(2y2+x2)=dxd​(1)
​

We know what the derivative of ​x2​ is:

​
(x2)′=2x
​

And ​dxd​(1)=0​ because ​1​ is a constant.


But what about ​2y2? Well since ​y​ depends on ​x, we use the chain rule:

​
dxd​(2y2)=2⋅dyd(y2)​⋅dxdy​
​

But ​dyd(y2)​=2y, so

​
dxd​(2y2)=4y⋅dxdy​
​

Plugging everything back in gives:

​
4y⋅dxdy​+2x=0
​

We can rearrange to find

​
dxdy​=4y2x​=yx​
​


This is an example of a technique called implicit differentiation. It probably seems super weird right now, but after a few practice examples this will start to feel routine.


You might notice that in this case ​dxdy​=yx​, which depends on ​x​ and ​y. This is often the case in implicit differentiation, simply because the curve is not a pure function of ​x, so the slope is usually not either (I say usually because sometimes the curves are symmetric).