What is the derivative of (sinx)^(cos^(-1)x)?
2 Answers
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) )
Explanation:
Let,
Here, by the Chain Rule,
For the Derivative on the R.H.S., we use the Product Rule
and the Chain Rule.
Using
Enjoy Maths.!