What is the derivative of (sinx)^(cos^(-1)x)?

2 Answers
Jul 4, 2017

dy/dx = (sinx)^(cos^(-1)x) \ (cos^(-1)x \ cotx -(ln sinx)/sqrt(1-x^2) )

Explanation:

Let:

y=(sinx)^(cos^(-1)x)

Take Natural logarithms of both sides:

ln y = ln {(sinx)^(cos^(-1)x) }
" " = (cos^(-1)x) \ ln {sinx}

Differentiate Implicitly,applying the product rule and chain rule:

1/y \ dy/dx = (cos^(-1)x)(d/dx ln {sinx}) + (d/dx cos^(-1)x)(ln {sinx})
" " = (cos^(-1)x)(1/sinx \ cosx) + (-1/sqrt(1-x^2))(ln {sinx})
" " = cos^(-1)x \ cosx/sinx -1/sqrt(1-x^2)(ln {sinx})
" " = cos^(-1)x \ cotx -(ln sinx)/sqrt(1-x^2)

And so:

dy/dx = y \ (cos^(-1)x \ cotx -(ln sinx)/sqrt(1-x^2) )
" " = (sinx)^(cos^(-1)x) \ (cos^(-1)x \ cotx -(ln sinx)/sqrt(1-x^2) )

Jul 4, 2017

[(sinx)^(cos^(-1)x){sqrt(1-x^2)cotxcos^(-1)x-ln(sinx)}]/sqrt(1-x^2).

Explanation:

Let, y=(sinx)^(cos^(-1)x).

:. lny=ln{(sinx)^(cos^-1x)}=(cos^(-1)x)(ln(sinx)).

:. d/dx(lny)=d/dx{(cos^(-1)x)(ln(sinx))}............(ast).

Here, by the Chain Rule,

d/dx(lny)=d/dy(lny)*d/dx(y)=1/y*dy/dx......(ast^1).

For the Derivative on the R.H.S., we use the Product Rule

and the Chain Rule.

d/dx{(cos^(-1)x)(ln(sinx))},

=cos^(-1)x*d/dx(ln(sinx))+ln(sinx)*d/dxcos^(-1)x,

=(cos^(-1)x){1/sinx*d/dx(sinx)}
+ln(sinx)*{-1/sqrt(1-x^2)},

=(cos^(-1)x)(cosx/sinx)-(ln(sinx))/sqrt(1-x^2),

=cotxcos^(-1)x-ln(sinx)/sqrt(1-x^2),

={sqrt(1-x^2)cotxcos^(-1)x-ln(sinx)}/sqrt(1-x^2).....(ast^2).

Using (ast^1) and (ast^2)" in "(ast), we have,

1/y*dy/dx={sqrt(1-x^2)cotxcos^(-1)x-ln(sinx)}/sqrt(1-x^2),

rArr dy/dx=[y{sqrt(1-x^2)cotxcos^(-1)x-ln(sinx)}]/sqrt(1-x^2), or,

dy/dx=[(sinx)^(cos^(-1)x){sqrt(1-x^2)cotxcos^(-1)x-ln(sinx)}]/sqrt(1-x^2).

Enjoy Maths.!