Question

derivada de y=sen^3 4x

Answer 1

$\mathrm{y=\sin^3{4x}\ \ \| \ \ y'=\frac{d}{dx}\big{[\sin^3{\mathrm{4x}}\big]}}\\\\ \textrm{Aplicando a regra de pot\^encia, teremos que:}\\\\ \mathrm{\frac{d}{dx}{[u(x)^n]}=n.u(x)^{n-1}.u'(x)\ \to\ y'=3\sin^2{4x}.\frac{d}{dx}{[\sin{4x}]}}\\\\ \textrm{Pela regra da cadeia, teremos que:}\\\\ \mathrm{\frac{d}{dx}{[f(g(x))]}=f'(g(x)).g'(x)\ \to\ y'=3\sin^2{4x}.\cos{4x}.\frac{d}{dx}{[4x]}}\\\\ \mathrm{y'=3\sin^2{4x}.\cos{4x}.4\ \to\ \boxed{\mathbf{y'=12\cos{4x}\sin^2{4x}}}}$