Question

How do you differentiate `3sin^5(2x^2)` using the chain rule?

Answer 1

Break the given chain rule problem in manageable links and use it to differentiate. Explanation is given below.

Explanation:

The Chain rule is easy if you split the chain into manageable links.

Let us see how we can do this with respect to our problem

`3sin^5(2x^2)`
`y=3sin^5(2x^2)`

We can split the links as below
`y=3u^5`
`u=sin(v)`
`v=2x^`

The chain rule is given by the following
`dy/dx = dy/(du)* (du)/(dv) * (dv)/(dx)`

Note how each link is differentiated separately and multiplied to form the chain.

`y=3u^5`
`dy/(du) = 15u^4`

`u=sin(v)`
`(du)/(dv) = cos(v)`

`v=2x^2`
`(dv)/(dx) = 4x`

Applying chain rule we get

`dy/dx = 15u^4*cos(v)*4x`
`dy/dx = 60xu^4cos(v)`
`dy/dx=60xsin^4(v)cos(v)` substituted `u=sin(v)`
`dy/dx = 60xsin^4(2x^2)cos(2x^2)` substituted `v=2x^2`

Final answer

`dy/dx = 60xsin^4(2x^2)cos(2x^2)`

To master try few problems and see how to split the links and differentiate. Practice would help you do similar problems with ease.