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Question

Find the derivative of $y = \frac{1}{8} {(2 + \cos^{3} (6 t))}^{4}$ .

Accepted Answer

$y^{'} = - 9 \cos^{2} (6 t) \sin (6 t) {(2 + \cos^{3} (6 t))}^{3}$

Answer

$y^{'} = - 9 \cos^{2} (6 t) \sin (6 t) {(2 + \cos^{3} (6 t))}^{3}$

Answer

$y^{'} = - 9 \cos^{2} (6 t) \sin (6 t) {(2 + \cos^{3} (6 t))}^{3}$

Answer

$y^{'} = - 9 \cos^{2} (6 t) \sin (6 t) {(2 + \cos^{3} (6 t))}^{3}$

Find the derivative of y=18(2+cos3(6t))4y=\frac18\left(2+\cos^3\left(6t\right)\right)^4.