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:

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

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

Now collecting ​dxdy​​ terms:

So

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​.

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.

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).

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

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:

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

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

Now collecting ​dxdy​​ terms:

So

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​.

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.

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).

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