Question

Alguém pode me ajudar nesse exercício?
O assunto é sobre função derivada usando a regra do quociente

Answer 1

A regra do quociente diz que a derivada do quociente entre duas funções, f(x) e g(x), pode ser calculada por:

$\boxed{\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)~=~\dfrac{\frac{df(x)}{dx}\cdot g(x)~-~\frac{dg(x)}{dx}\cdot f(x)}{\left(\,g(x)\,\right)^2}}\\\\\\Ou,~seguindo~uma~notacao~mais~simples~pra~derivada:\\\\\\\boxed{\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)~=~\dfrac{f'(x)\cdot g(x)~-~g'(x)\cdot f(x)}{\left(\,g(x)\,\right)^2}}$

Sendo f(x)=5x³+4x e g(x)=x², temos:

$y'(x)~=~\dfrac{(5x^3+4x)'\cdot x^2~-~(x^2)'\cdot(5x^3+4x)}{\left(x^2\right)^2}\\\\\\y'(x)~=~\dfrac{(3\cdot5x^{3-1}+1\cdot4x^{1-1})\cdot x^2~-~(2\cdotx^{2-1})\cdot(5x^3+4x)}{x^4}\\\\\\y'(x)~=~\dfrac{(15x^{2}+4x^{0})\cdot x^2~-~(2x^{1})\cdot(5x^3+4x)}{x^4}\\\\\\y'(x)~=~\dfrac{(15x^{2}+4)\cdot x^2~-~(2x)\cdot(5x^3+4x)}{x^4}\\\\\\y'(x)~=~\dfrac{(15x^2\cdot x^2~+~4\cdot x^2)~-~(2x\cdot5x^3~+~2x\cdot4x)}{x^4}\\\\\\y'(x)~=~\dfrac{(15x^4~+~4x^2)~-~(10x^4~+~8x^2)}{x^4}$

$y'(x)~=~\dfrac{15x^4~+~4x^2~-~10x^4~-~8x^2}{x^4}\\\\\\y'(x)~=~\dfrac{5x^4~-~4x^2}{x^4}\\\\\\y'(x)~=~\dfrac{x^2\cdot(5x^2~-~4)}{x^4}\\\\\\y'(x)~=~\dfrac{x^{2\!\!\!\backslash}\cdot(5x^2~-~4)}{x^{^24\!\!\!\backslash}}\\\\\\\boxed{y'(x)~=~\dfrac{5x^2~-~4}{x^2}}\\\\\\\\\Huge{\begin{array}{c}\Delta \tt{\!\!\!\!\!\!\,\,o}\!\!\!\!\!\!\!\!\:\,\perp\end{array}}Qualquer~d\acute{u}vida,~deixe~ um~coment\acute{a}rio$