What is the derivative of ${sin}^{2} (x)$ $sin^2(x)$ ?

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Answer 1 · 2016-01-09T06:40:43

The derivative is $2 sin (x) cos (x)$ $2sin(x)cos(x)$ or it can also be written as $sin (2 x)$ $sin(2x)$

$y = {sin}^{2} (x)$ $y=sin^2(x)$

Let us use the chain rule.

Let $y = u^{2}$ $y=u^2$ and $u = sin (x)$ $u = sin(x)$

Chain rule:

$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $dy/dx = dy/(du)xx(du)/dx$

$y = u^{2}$ $y=u^2$

$\frac{d y}{d u} = 2 u$ $dy/(du) = 2u$

$u = sin (x)$ $u=sin(x)$

$\frac{d u}{d x} = cos (x)$ $(du)/dx = cos(x)$

$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $dy/dx = dy/(du)xx(du)/dx$

$\frac{d y}{d x} = 2 u \times cos (x)$ $dy/dx = 2uxxcos(x)$

$\frac{d y}{d x} = 2 sin (x) cos (x)$ $dy/dx = 2sin(x)cos(x)$

$\frac{d y}{d x} = sin (2 x)$ $dy/dx = sin(2x)$

What is the derivative of sin^2(x)sin2(x)sin^2(x)?