Dy/dx? Ln(sin^-1(x))

1 Answer
Apr 24, 2018

d/dxln(sin^-1x)=1/(sin^-1xsqrt(1-x^2))

Explanation:

The Chain Rule, when applied to logarithms, tells us that if u is some function in terms of x, then

d/dx(lnu)=1/u*(du/dx)

For ln(sin^-1x), we see:

u=sin^-1x
(du)/dx=1/sqrt(1-x^2)

So,

d/dxln(sin^-1x)=1/sin^-1x*1/sqrt(1-x^2)

d/dxln(sin^-1x)=1/(sin^-1xsqrt(1-x^2))