Question

Find the second derivative of y=sinsquarex​

Answer 1

Answer:

y=sin^2x

dy/dx = 2sinxcosx

y'=2sinxcosx

y"=2{sinx(-sinx)+cosx(cosx)}

y"=2{-sin^2x+cos^2x}

Answer 2

Answer:

2cos2x

Step-by-step explanation:

$y = \sin^{2} x \\ differenciating \: with \: respect \: to \\ x \: we \: get \\ \frac{dy}{dx} = 2 \sin(x) \cos(x) \\ \frac{dy}{dx} = \sin(2x) \\ again \: differenciating \: with \: respect \: to \: x \\ { \frac{dy}{{dx}^{2} }}^{2} = 2 \cos(2x)$