How do you find the derivative of $y = {sin}^{2} x$ $y = sin^2 x$ ?

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Answer 1 · 2018-05-09T08:33:17

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

Using $u = sin x$ $u=sinx$ gives us $y = u^{2}$ $y=u^2$

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

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

$\frac{d y}{d x} = 2 u cos x = 2 sin x cos x$ $dy/dx=2ucosx=2sinxcosx$

How do you find the derivative of y = sin^2 xy=sin2x y = sin^2 x?