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

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Answer 1 · 2016-06-04T03:44:33

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

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.

How do I find the derivative of y = (sin(e^x))^ln(x^2)y=(sin(ex))ln(x2)y = (sin(e^x))^ln(x^2)?