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

1 Answer
Karthik G
Dec 25, 2015

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.

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