How do you use the chain rule to differentiate $y = {sin}^{3} (2 x + 1)$ $y=sin^3(2x+1)$ ?

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Answer 1 · 2016-08-08T14:20:44

$\frac{d y}{d x} = 6 {sin}^{2} (2 x + 1) cos (2 x + 1)$ $(dy)/(dx) = 6sin^2(2x+1)cos(2x+1)$

$u (x) = 2 x + 1$ $u(x) = 2x+1$ so $\frac{d u}{d x} = 2$ $(du)/(dx) = 2$

$y = {sin}^{3} (u) \Rightarrow \frac{d y}{d u} = 3 {sin}^{2} (u) cos (u)$ $y = sin^3(u) implies (dy)/(du) = 3sin^2(u)cos(u)$

$\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ $(dy)/(dx) = (dy)/(du)(du)/(dx)$

$\frac{d y}{d x} = 6 {sin}^{2} (2 x + 1) cos (2 x + 1)$ $(dy)/(dx) = 6sin^2(2x+1)cos(2x+1)$

How do you use the chain rule to differentiate y=sin^3(2x+1)y=sin3(2x+1)y=sin^3(2x+1)?