How do you differentiate f(x)=(x-sin3x)^3/9 using the chain rule?

Jan 9, 2018

$\frac{1}{3} \cdot \left(1 - 3 \cos 3 x\right) \cdot {\left(x - \sin 3 x\right)}^{2}$

Explanation:

$\frac{d}{\mathrm{dy}} \left(\frac{{\left(x - \sin 3 x\right)}^{3}}{9}\right)$

= $\frac{d}{\mathrm{dy}} \left(\frac{1}{9} {\left(x - \sin 3 x\right)}^{3}\right)$

= $\frac{1}{9} \cdot 3 \cdot \left(1 - \cos 3 x \cdot 3\right) \cdot {\left(x - \sin 3 x\right)}^{2}$

= $\frac{1}{3} \cdot \left(1 - 3 \cos 3 x\right) \cdot {\left(x - \sin 3 x\right)}^{2}$