What is the derivative of (lnx)^(sinx)?

1 Answer
Jan 9, 2016

dy/dx = (ln(x))^sin(x){sin(x)/(xln(x))+cos(x)ln(ln(x))}

Explanation:

Let y=(ln(x))^sin(x)

To find the derivative of such a problem we need to take logarithms on both the sides.

ln(y)=ln((ln(x))^sin(x))

This step is done to move the exponent to the front of the equation.
as color(brown)(ln(a^n) = n*ln(a)

ln(y)=sin(x)*ln(ln(x))

Now let us differentiate both sides with respect to x

We shall use product rule color(blue)((uv)'=uv'+vu'

(1/y)dy/dx = sin(x)d/dx(ln(ln(x))) + ln(ln(x))d/dx(sin(x)

color(blue)"Derivative of ln(ln(x)) to be done using chain rule"

(1/y)dy/dx = sin(x)(1/ln(x))*d/dx(ln(x)+ln(ln(x))(cos(x))

(1/y)dy/dx =sin(x)(1/ln(x))*1/x+cos(x)ln(ln(x))

(1/y)dy/dx = sin(x)/(xln(x))+cos(x)ln(ln(x))

dy/dx = y{sin(x)/(xln(x))+cos(x)ln(ln(x))}

dy/dx = (ln(x))^sin(x){sin(x)/(xln(x))+cos(x)ln(ln(x))}