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

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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)#