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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
Review VideosFormula BookletMy Progress
BlogLanding Page
Sign UpLogin
Perplex
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Differentiation
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Related Rates
n^th Derivative
Related Rates
Differentiation

Related Rates

0 of 0 exercises completed
Using implicit differentiation and the chain rule to relate changing quantities, including \(\frac{\differentialD y}{\differentialD t}=\frac{\differentialD y}{\differentialD x}\cdot\frac{\differentialD x}{\differentialD t}\), and applying this to volume, distance, and angle rates.

Want a deeper conceptual understanding? Try our interactive lesson! (Plus Only)

Related Rates
AHL 5.14

Given three variables ​x,  ​y, and ​z,

​
dzdy​=dzdx​⋅dxdy​.
​


Hence, given ​dzdx​,​ we can find an expression for ​dzdy​​ by calculating ​dxdy​.

Implicit Differentiation
AHL 5.14

Implicit differentiation is when we differentiate both sides of an equation. It is helpful when we have an equation that cannot be simplified to ​y=f(x).


For example:

​
xy2=x+y
​

Differentiating both sides with respect to ​x​ and using the product rule:

​
(x)′⋅y2+x⋅(y2)′=x′+y′
​

By the chain rule, we know that ​(y2)′=2y⋅dxdy​:

​
1⋅y2+2xy⋅dxdy​=1+dxdy​
​

Now collecting ​dxdy​​ terms:

​
y2−1=dxdy​(1−2xy)
​

So

​
dxdy​=1−2xyy2−1​
​
Related rates with implicit differentiation
AHL 5.14

Since ​dtdy​=dxdy​⋅dtdx​, you may be asked to use implicit differentiation to find ​dxdy​, then with a given ​dtdx​​ and point, you can find ​dtdy​.

Volume related rates
AHL 5.14

Given the time rate of change of radius, length, height, or width of a three dimensional object, you may find the time rate of change of volume by taking the derivative of the volume equation.

Distance related rates
AHL 5.14

Let ​L​ be the distance from the origin of a point with coordinates ​(x,y). Then, given ​dtdx​​ and ​dtdy​, we can find ​dtdL​​ at a given point ​(x,y).

Angle related rates
AHL 5.14

Using a given rate of change ​dtdx​​ and trigonometry, we can calculate ​dxdθ​, which can be used to find ​dtdθ​.​

Nice work completing Related Rates, here's a quick recap of what we covered:

Skills covered

Mixed Practice

Exercises checked off

I'm Plex, here to help you understand this concept!
/
Differentiation
/
Related Rates
n^th Derivative
Related Rates
Differentiation

Related Rates

0 of 0 exercises completed
Using implicit differentiation and the chain rule to relate changing quantities, including \(\frac{\differentialD y}{\differentialD t}=\frac{\differentialD y}{\differentialD x}\cdot\frac{\differentialD x}{\differentialD t}\), and applying this to volume, distance, and angle rates.

Want a deeper conceptual understanding? Try our interactive lesson! (Plus Only)

Related Rates
AHL 5.14

Given three variables ​x,  ​y, and ​z,

​
dzdy​=dzdx​⋅dxdy​.
​


Hence, given ​dzdx​,​ we can find an expression for ​dzdy​​ by calculating ​dxdy​.

Implicit Differentiation
AHL 5.14

Implicit differentiation is when we differentiate both sides of an equation. It is helpful when we have an equation that cannot be simplified to ​y=f(x).


For example:

​
xy2=x+y
​

Differentiating both sides with respect to ​x​ and using the product rule:

​
(x)′⋅y2+x⋅(y2)′=x′+y′
​

By the chain rule, we know that ​(y2)′=2y⋅dxdy​:

​
1⋅y2+2xy⋅dxdy​=1+dxdy​
​

Now collecting ​dxdy​​ terms:

​
y2−1=dxdy​(1−2xy)
​

So

​
dxdy​=1−2xyy2−1​
​
Related rates with implicit differentiation
AHL 5.14

Since ​dtdy​=dxdy​⋅dtdx​, you may be asked to use implicit differentiation to find ​dxdy​, then with a given ​dtdx​​ and point, you can find ​dtdy​.

Volume related rates
AHL 5.14

Given the time rate of change of radius, length, height, or width of a three dimensional object, you may find the time rate of change of volume by taking the derivative of the volume equation.

Distance related rates
AHL 5.14

Let ​L​ be the distance from the origin of a point with coordinates ​(x,y). Then, given ​dtdx​​ and ​dtdy​, we can find ​dtdL​​ at a given point ​(x,y).

Angle related rates
AHL 5.14

Using a given rate of change ​dtdx​​ and trigonometry, we can calculate ​dxdθ​, which can be used to find ​dtdθ​.​

Nice work completing Related Rates, here's a quick recap of what we covered:

Skills covered

Mixed Practice

Exercises checked off

I'm Plex, here to help you understand this concept!

Generating starter questions...

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Generating starter questions...

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