If $y = {cos x}^{3} {sin}^{2} (x^{5})$ $y = cosx^3sin^2(x^5)$ , what is $\frac{d y}{d x}$ $dy/dx$ ?

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Answer 1 · 2017-11-19T01:40:01

See explanation

According to the product rule,
$d/dx[f(x)*g(x)]=f(x)*g'(x)+g(x)*f'(x)$ , where $f(x) = cos(x^3)$ and $g(x) = sin^2 (x^5)$

Therefore (be careful with the chain rule),
$d/dx [cosx^3 *sin^2 (x^5)] = cosx^3 * 2sin(x^5) * cos(x^5) * 5x^4 +sin^2(x^5) * -sin(x^3)*3x^2$

That can be re-written or simplified as:
$x^2sin(x^5)(10x^2cos(x^3)cos(x^5)-3sin(x^3)sin(x^5))$

If y = cosx^3sin^2(x^5)y=cosx3sin2(x5)y = cosx^3sin^2(x^5), what is dy/dxdydxdy/dx?