How do you find the derivative of #[x+(x+sin^2x)^3]^4#?

Question

17250 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-07T06:55:45

#d/dx[x+(x+sin^2 x)^3]^4#
#=4[x+(x+sin^2 x)^3]^3*[1+3(x+sin^2 x)^2*(1+sin 2x)]#
Take note: #sin 2x=2*sin x*cos x#

From the given

#[x+(x+sin^2 x)^3]^4#

#d/dx[x+(x+sin^2 x)^3]^4=#
#4*[x+(x+sin^2 x)^3]^(4-1)*d/dx(x+(x+sin^2 x)^3)#

#=4*[x+(x+sin^2 x)^3]^3*(1+3*(x+sin^2 x)^(3-1)*d/dx(x+sin^2 x))#

#=4*[x+(x+sin^2 x)^3]^3*(1+3*(x+sin^2 x)^2*(1+2*sin x*cos x))#

#=4*[x+(x+sin^2 x)^3]^3*(1+3*(x+sin^2 x)^2*(1+sin 2x))#

God bless....I hope the explanation is useful.