Question

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

Answer 1

`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.