How do you find the derivative of $y=(sinx)^(x^3)$ ?

Question

11703 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-12T07:40:17

$(dy)/(dx)=x^2(sinx)^(x^3)(x^2lnsinx+xcotx)$

$y=(sinx)^(x^3)$

take natural logs

$lny=ln(sinx)^(x^3)$

$=>lny=x^3lnsinx$

$d/(dx)(lny)=d/(dx)(x^3lnsinx)$

differentiate wrt $x$

RHS using the product rule

$1/y(dy)/(dx)=3x^2lnsinx+x^3cosx/sinx$

$(dy)/(dx)=y[x^2(3lnsinx+xcotx)]$

$(dy)/(dx)=x^2(sinx)^(x^3)(x^2lnsinx+xcotx)$

How do you find the derivative of y=(sinx)^(x^3)y=(sinx)^(x^3)?