## Question

###### How do I find the derivative of #y = (sin(e^x))^ln(x^2)#?

## Answers

#color(blue)(y'=(sin e^x)^(ln x^2)[e^x*ln x^2*cot e^x+2/x*ln sin e^x])# ## Explanation:

The given equation is

#y=(sin e^x)^(ln x^2)# Take the natural logarithm of both sides then differentiate with respect to x

#y=(sin e^x)^(ln x^2)#

#ln y=ln (sin e^x)^(ln x^2)#

#ln y=(ln x^2)*ln (sin e^x)# We now have a derivative of a product at the right sides of the equation.

Obtain the derivative of both sides

#ln y=(ln x^2)*ln (sin e^x)#

#d/dx(ln y)=d/dx[(ln x^2)*ln (sin e^x)]#

#1/y*y'=(ln x^2)*d/dx(ln sin e^x)+(ln sin e^x)*d/dx(ln x^2)#

#1/y*y'=(ln x^2)*1/( sin e^x)*d/dx(sin e^x)+(ln sin e^x)*1/x^2*d/dx(x^2)#

#1/y*y'=(ln x^2)*1/( sin e^x)(cos e^x)d/dx(e^x)+(ln sin e^x)*1/x^2*2x#

#1/y*y'=(ln x^2)*1/( sin e^x)(cos e^x)(e^x)+2/x*ln sin e^x#

#1/y*y'=(ln x^2)*(cot e^x)(e^x)+2/x*ln sin e^x# Multiply both sides by

#y# then simplify

#y*1/y*y'=y*(ln x^2)*(cot e^x)(e^x)+2/x*ln sin e^x#

#y'=y*(ln x^2)*(cot e^x)(e^x)+2/x*ln sin e^x#

#y'=(sin e^x)^(ln x^2)[e^x*ln x^2*cot e^x+2/x*ln sin e^x]# God bless....I hope the explanation is useful.