differentiate tan^-1(x-1/x+1) with respect to sin^-1(3x-4x^3), if -1/2

Y=tan^-1((x-1)/(x+1))

Dy/dx=[1/{1+((x-1)/(x+1))²}].[(x+1).1-(x-1).1]/(x+1)²

=[1/{(x+1)²+(x-1)²}].2

=2/2(x²+1)=1/(1+x²)

Let z=sin^-1(3x-4x³)

=3sin^-1x

Dz/dx=3/v(1-x²)

Then dy/dz=(dy/dx)/(dz/dx)={1/(1+x²)}/{3/v(1-x²)}

=v(1-x²)/3(1+x²) Ans.

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Rinkoo Gupta

AskIITians Faculty