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

1 Answer
Jim H
Feb 5, 2016

See the explanation below.

Explanation:

If we leave it as written , we have:

#(sin^3(2x^2) ) ^(1/2)#

So the derivative is

#1/2 (sin^3(2x^2) )^(-1/2)[d/dx(sin^3(2x^2)]#

# = 1/2 (sin^3(2x^2) )^(-1/2)[3sin^2(2x^2)(d/dx(sin(2x^2)))]#

# = 1/2 (sin^3(2x^2) )^(-1/2)[3sin^2(2x^2)cos(2x^2)(d/dx(2x^2))]#

# = 1/2 (sin^3(2x^2) )^(-1/2)[3sin^2(2x^2)cos(2x^2)(4x)]#

Now simplify.

If we rewrite as

#(sin(2x^2) ) ^(3/2)#

We get

#3/2 (sin(2x^2))^(1/2) cos(2x^2) (4x)#

# = 6xcos(2x^2)sqrt(sin(2x^2))#

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