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Perplex
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Differentiation
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Implicit differentiation
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Implicit differentiation
Differentiation

Implicit differentiation

0 of 0 exercises completed

Differentiating both sides of an equation with respect to ​x​ to find ​dxdy​​ for relations not written as ​y=f(x), using the product and chain rules, and finding horizontal and vertical tangents and normals from the resulting derivative.

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​
​
Implicit Derivative: Horizontal and vertical tangents & normals
AHL 5.14

In problems involving implicit derivatives, you may be asked to solve for points where the tangent to the curve is horizontal or vertical. A horizontal tangent means ​dxdy​=0, and a vertical tangent occurs in the case where ​dxdy​=denominatornumerator​​ and the denominator equals zero.

If the question asks for vertical / horizontal normals, just recall that a vertical normal means a horizontal tangent, and vice-versa.

Nice work completing Implicit differentiation, 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
/
Implicit differentiation
Related Rates
Implicit differentiation
Differentiation

Implicit differentiation

0 of 0 exercises completed

Differentiating both sides of an equation with respect to ​x​ to find ​dxdy​​ for relations not written as ​y=f(x), using the product and chain rules, and finding horizontal and vertical tangents and normals from the resulting derivative.

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​
​
Implicit Derivative: Horizontal and vertical tangents & normals
AHL 5.14

In problems involving implicit derivatives, you may be asked to solve for points where the tangent to the curve is horizontal or vertical. A horizontal tangent means ​dxdy​=0, and a vertical tangent occurs in the case where ​dxdy​=denominatornumerator​​ and the denominator equals zero.

If the question asks for vertical / horizontal normals, just recall that a vertical normal means a horizontal tangent, and vice-versa.

Nice work completing Implicit differentiation, 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!

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