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

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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.

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