Question

Calcule a derivada
f(x)= sen³3x

Answer 1

$\displaystyle i)~~~~f(x)=\sin^3(3x)\\\\ii)~~~\frac{d}{dx}(\sin^3(3x))\implies \sin (3x)=u~~~~3x=v\\\\iii)~~\frac{d}{dx}\sin^3(3x)=\frac{d}{du}u^3\frac{du}{dv}\cdot\frac{dv}{dx}\\\\iv)~~\frac{d}{dx}\sin^33x=3u^2\cdot \cos(v)\cdot 3\\\\v)~~~\frac{d}{dx}\sin^33x=9u^2\cos(v)\implies u=\sin(v)\implies v=3x\\\\vi)~~\frac{d}{dx}\sin^33x=\boxed{\boxed{9\sin^2(3x)\cos(3x)}}$
Ou na notação de f':
$\displaystyle i)~~~~f(3x)=\sin^3(3x)\\\\ii)~~~g(h(x))=\sin(h(x))\implies h(x)=3x\\\\iii)~~f(g)=g^3\implies f(g(h(x))=g^3(h(x))\implies f(g(h(x)))=\sin^3(3x)\\\\iv)~~~f'(g(h(x)))=f'(g(h(x)))\cdot g'(h(x))\cdot h'(x)\\\\v)~~~f'(g(h(x)))=3g^2(h(x))~~~~~~~~~~g'(h(x))=\cos(h(x))~~~~~h'(x)=3\\\text{logo:}\\\\vi)~~~f'(g(h(x)))=3(\cos^2(3x))\cdot 3=\boxed{9\cos^2(3x)}$

observação:
No passo iii) utilizei a regra da cadeia:
$\boxed{\frac{d}{dx}f(g(x))=\frac{df}{dg}\cdot\frac{dg}{dx}}\implies \boxed{f'(g(x))=f'(g(x))\cdot g'(x)}$

Recomendo estudar a notação de Leibnitz para derivadas, ajudam bastante a compreender cálculo integral através do diferencial, principalmente quando você for estudar integral.

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