How do you find the derivative of #y=cosln(4x^3)#?

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Answer 1 · 2017-02-05T03:07:56

#(dy)/(dx)=-(3sin(ln(4x^3)))/x#

Using chain rule, let #u=ln(4x^3)#

So #y=cosu#

#(dy)/(dx)=(dy)/(du)*(du)/(dx)#

#=-sinu*d/(dx)ln(4x^3)#

Now we have to derive #ln(4x^3)# using chain rule as well.

Let #y=lnu, u=4x^3#

#(dy)/(dx)=d/(du)lnu*d/(dx)4x^3#

#=1/u*12x^2#

#=(12x^2)/(4x^3)#

#=3/x#

#:. d/(dx)ln(4x^3)=-sin(ln(4x^3))*3/x#

#=-(3sin(ln(4x^3)))/x#