What is the derivative of #y=(sinx)^(2x)#?

Question

5229 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-02T12:14:14

#dy/dx = 2sin^(2x)(x)(x cotx + ln(sinx))#

#y = (sinx)^(2x) = sin^(2x)(x)#

#lny =2x*ln(sinx)#

Applying Implicit differentiation, the product rule, standard differentials and the chain rule:

#1/y dy/dx = 2x*1/sinx * cosx + ln(sinx)*2#

#dy/dx = y*(2x*cosx/sinx +2ln(sinx))#

#= y* 2(xcotx+ln(sinx))#

#= (sinx)^(2x)*2(xcotx+ln(sinx))#

#= 2sin^(2x)(x)(x cotx + ln(sinx))#