Question

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

Answer 1

`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) )`

Answer 2

`[(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.!