Question

Questão de derivada ​

Answer 1

(item a)

$\displaystyle y' = [Cos(\sqrt{Sen(Tg(\pi.x))} \ )]'$

aplicando a regra da cadeia:

$\displaystyle y' = Cos'(\sqrt{Sen(Tg(\pi.x))}).[Sen(Tg(\pi.x))]^{\displaystyle \frac{1}{2}}]'$

$\displaystyle y' = -Sen(\sqrt{Sen(Tg(\pi.x))}).\frac{1}{2}.[Sen(Tg(\pi.x))]^{\displaystyle (\frac{1}{2}-1)}.[Sen(Tg(\pi.x))]'$

$\displaystyle y' = -Sen(\sqrt{Sen(Tg(\pi.x))}).\frac{1}{2.\sqrt{Sen(Tg(\pi.x))}}.[Sen(Tg(\pi.x))]'$

$\displaystyle y' = .\frac{-Sen(\sqrt{Sen(Tg(\pi.x))})}{\ \ \ \ \ \ 2.\sqrt{Sen(Tg(\pi.x))}}.Cos(Tg(\pi.x)).[Tg(\pi.x)]'$

$\displaystyle y' = .\frac{-Sen(\sqrt{Sen(Tg(\pi.x))}).Cos(Tg(\pi.x)).Sec^2(\pi.x)}{\ \ \ \ \ \ 2.\sqrt{Sen(Tg(\pi.x))}}.[\pi.x]'$

$\displaystyle y' = .\frac{-Sen(\sqrt{Sen(Tg(\pi.x))}).Cos(Tg(\pi.x)).Sec^2(\pi.x).\pi }{\ \ \ \ \ \ 2.\sqrt{Sen(Tg(\pi.x))}}$

Agora só arrumando :

$\fbox{\displaystyle y' = \frac{- \pi.Sec^2(\pi.x).Cos(Tg(\pi.x)).Sen(\sqrt{Sen(Tg(\pi.x))}).}{\ \ \ \ \ \ 2.\sqrt{Sen(Tg(\pi.x))}} }$

Pronto está derivado.

(item b)

$\displaystyle y' = [\frac{(r^3-1)^5}{(2r+1)^3}]'$

Usando a regra do quociente :

$\displaystyle y' = [\frac{[(r^3-1)^5]'.(2+1)^3-(r^3-1)^5.[(2r+1)^3]']}{[(2r+1)^3]^2}]'$

Vamos derivar aqui e depois só substituir :

$\displaystyle [(r^3-1)^5]' = 5.(r^3-1)^{4}.(r^3-1)'$

$\displaystyle [(r^3-1)^5]' = 5.(r^3-1)^{4}.3r^2 \to \fbox{15.r^2.(r^3-1)^{4} }$

Agora vamos derivar o outro que falta :

$[(2r+1)^3]' = 3.(2r+1)^2.(2r+1)' \to \fbox{\displaystyle 6.(2r+1)^2 }$

e no denominador fica :

${(2x+1)^3]^2 = (2r+1)^6$

Substituindo :

$\displaystyle y' = \frac{15r^2(r^3-1)^4.(2r+1)^3-(r^3-1)^5.6r(2r+1)^2}{(2r+1)^6}$

Colocando o $(r^3-1)^4.(2r+1)^2$ em vidência :

$\fbox{\displaystyle y' = \frac{(r^3-1)^4.(2r+1)^2.[\ 15.r^2.(2r+1)-(r^3-1).6r \ ]}{(2r+1)^6} }$

Pronto, está derivado.

Qualquer dúvida é só falar.