How do you differentiate $y = sin 5 x - \frac{1}{3} {sin}^{3} 5 x$ $y = sin 5x - 1/3 sin^3 5x$ ?

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Answer 1 · 2018-03-25T12:27:30

$color(brown)(y' = 5 cos (5x) (1 - sin^2 (5x))$

$y = sin 5x - (1/3) sin^3 (5x)$

$y' = (dy)/(dx) = cos 5x * 5 - (1/cancel3) cancel3 sin^2 (5x) cos 5x * 5$

$y' = 5 cos (5x) - 5 sin^2 (5x) * cos (5x)$

$y' = 5 cos (5x) (1 - sin^2 (5x))$

How do you differentiate y = sin 5x - 1/3 sin^3 5xy=sin5x−13sin35x y = sin 5x - 1/3 sin^3 5x?