Question

find derivative of y=sin^3x+sin3x

Answer 1

$<b>Step-by-step explanation:</b>$

$Given:y = sin^3xsin3x$

$=\frac{d}{dx}(sin^3x)sin3x + \frac{d}{dx}(sin3x)sin^3x$

$=3sin^2x.cosxsin3x + cos3x.3sin^3x$

$=3sin^2x.cosxsin3x + cos3x.3sin^2xsinx$

$=3sin^2x.cosxsin3x + 3cos3x.sin^2xsinx$

$=3sin^2x[cosx.sin3x + cos3x.sinx]$

$=3sin^2x sin(x + 3x)$

$=\boxed{3sin^2x sin4x}$

Hope it helps!

Answer 2

3 sin^2x cosx+3cos 3x

Step-by-step explanation: