Question

Galera, me ajudem a derivar parcialmente em relação a x isso aqui por favor!!

$F= \frac{c \pi x^{2} }{4} \sqrt{y-z}$

Onde c é constante positiva

Pfvr galera tenho muita dificuldade em cvv, quem souber da um help aqui :)

Answer 1

Vamos derivar parcialmente, lembrando que:

$f(x)=g(x)\cdot h(x)\Longrightarrow \dfrac{\partial f(x)}{\partial x}=\dfrac{\partial g(x)}{\partial x}\cdot h(x)+\dfrac{\partial h(x)}{\partial x}\cdot g(x)$

Então:

$F=\dfrac{c\pi x^2}{4}\sqrt{y-z}\\\\ \dfrac{\partial F}{\partial x}=\dfrac{\partial}{\partial x}\left(\dfrac{c\pi x^2}{4}\sqrt{y-z}\right)\\\\ \dfrac{\partial F}{\partial x}=\dfrac{c\pi}{4}\cdot \dfrac{\partial}{\partial x}\left(x^2\cdot\sqrt{y-z}\right)\\\\ \dfrac{\partial F}{\partial x}=\dfrac{c\pi}{4}\cdot \left[\dfrac{\partial}{\partial x}(x^2)\cdot\sqrt{y-z}+x^2\cdot\dfrac{\partial}{\partial x}(\sqrt{y-z})\right]$

$\dfrac{\partial F}{\partial x}=\dfrac{c\pi}{4}\cdot [2x\cdot\sqrt{y-z}+x^2\cdot0]\\\\ \dfrac{\partial F}{\partial x}=\dfrac{c\pi}{4}\cdot [2x\cdot\sqrt{y-z}]\\\\ \boxed{\dfrac{\partial F}{\partial x}=\dfrac{c\pi x}{2}\sqrt{y-z}}$