Question

calcule as derivadas: u(x)=∛(sen^2 x)   e    v(x)=sen (cos x).

Answer 1

Regra da cadeia

$\boxed{\boxed{\dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)}}$
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a)

$u(x)=\sqrt[3]{sen^{2}(x)}=\sqrt[3]{(sen~x)^{2}}=(sen~x)^{2/3}$

Considere as seguintes funções:

$f(x)=x^{2/3}~~e~~g(x)=sen(x)$

Logo:

$f(g(x))=(g(x))^{2/3}=(sen~x)^{2/3}~~~\therefore~~~\boxed{\boxed{u(x)=f(g(x))}}$

Então, podemos derivar u pela regra da cadeia.

Antes disso, vamos achar as derivadas de f e g separadamente:

$f'(x)=\frac{d}{dx}x^{2/3}=(\frac{2}{3})x^{(2/3)-1}=(\frac{2}{3})x^{-1/3}\\\\g'(x)=\frac{d}{dx}sen(x)=cos(x)$

Logo, pela Regra da Cadeia:

$\dfrac{d}{dx}u(x)=f'(g(x))\cdot g'(x)\\\\\\u'(x)=\dfrac{2}{3}(g(x))^{-1/3}\cdot cos(x)\\\\\\u'(x)=\dfrac{2}{3}(sen~x)^{-1/3}\cdot cos(x)\\\\\\u'(x)=\dfrac{2}{3}\cdot\dfrac{1}{(sen~x)^{1/3}}\cdot cos(x)\\\\\\\boxed{\boxed{u'(x)=\dfrac{2}{3}\dfrac{cos(x)}{\sqrt[3]{sen(x)}}}}$

b)

$v(x)=sen(cos(x))$

Considere, agora, as funções

$f(x)=sen(x)~~e~~g(x)=cos(x)$

Então, podemos escrever v(x) como f(g(x)). Veja:

$f(g(x))=sen(g(x))=sen(cos(x))=v(x)$

Derivando f(x) e g(x):

$f'(x)=\frac{d}{dx}sen(x)=cos(x)\\\\g'(x)=\frac{d}{dx}cos(x)=-sen(x)$

Derivando v(x) pela regra da cadeia:

$v'(x)=f'(g(x))\cdot g'(x)\\\\v'(x)=cos(g(x))\cdot(-sen(x))\\\\v'(x)=cos(cos(x))\cdot(-sen(x))\\\\\boxed{\boxed{v'(x)=-sen(x)\cdot cos(cos(x))}}$