What is the derivative of $y=ln(x^4sin^2x)$ ?

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Answer 1 · 2016-07-14T18:38:05

$dy/dx=4/x+2cotx.$

We first simplify $y$ using the Rules of Log.

$y=ln(x^4*sin^2x)=lnx^4+lnsin^2x=4lnx+2lnsinx$

$:. dy/dx=(4lnx)'+(2lnsinx)'$

$=4*1/x+2*1/sinx*(sinx)'$ .................[Chain Rule]
$=4/x+2/sinx*cosx$

$:. dy/dx=4/x+2cotx.$

What is the derivative of y=ln(x^4sin^2x)y=ln(x^4sin^2x)?