Question

Desafio para matematicos:
Calcule a derivada das seguintes funções:
a) f(x)=(x-e^2x+1/x²)^4
b)g(x)=2x²-In(x+1/4-x)
Se possivel explicar a resolução para que eu possa aprender a fazer;

Answer 1

$\large\begin{array}{l} \textsf{(a) }\mathsf{f(x)=\left(\dfrac{x-e^{2x+1}}{x^2}\right)^{\!\!4}}\\\\ \textsf{Podemos enxergar f como uma fun\c{c}\~ao composta. Observe:}\\\\ \left\{\! \begin{array}{l} \mathsf{f(x)=u^4}\\\\ \mathsf{u=\dfrac{x-e^{2x+1}}{x^2}}\qquad\textsf{(aqui temos um quociente)} \end{array} \right. \end{array}$


$\large\begin{array}{l} \textsf{Ent\~ao, usamos a Regra da cadeia para derivar:}\\\\ \mathsf{\dfrac{df}{dx}=\dfrac{df}{du}\cdot \dfrac{du}{dx}}\\\\ \mathsf{\dfrac{df}{dx}=\dfrac{d}{du}(u^4)\cdot \dfrac{d}{dx}\!\left(\dfrac{x-e^{2x+1}}{x^2}\right)}\\\\ \mathsf{\dfrac{df}{dx}=4u^{4-1}\cdot \dfrac{\frac{d}{dx}(x-e^{2x+1})\cdot x^2-(x-e^{2x+1})\cdot \frac{d}{dx}(x^2)}{(x^2)^2}}\\\\ \mathsf{\dfrac{df}{dx}=4u^3\cdot \dfrac{\big(1-\frac{d}{dx}(e^{2x+1})\big)\cdot x^2-(x-e^{2x+1})\cdot 2x}{x^4}} \end{array}$

$\large\begin{array}{l} \mathsf{\dfrac{df}{dx}=4u^3\cdot \dfrac{\big(1-e^{2x+1}\cdot \frac{d}{dx}(2x+1)\big)\cdot x^2-(x-e^{2x+1})\cdot 2x}{x^4}}\\\\ \mathsf{\dfrac{df}{dx}=4u^3\cdot \dfrac{(1-e^{2x+1}\cdot 2)\cdot x^2-(x-e^{2x+1})\cdot 2x}{x^4}}\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{df}{dx}=4\left(\dfrac{x-e^{2x+1}}{x^2}\right)^{\!\!3}\cdot \dfrac{(1-2e^{2x+1})\cdot x^2-(x-e^{2x+1})\cdot 2x}{x^4}} \end{array}} \end{array}$

$\large\begin{array}{l} \textsf{Como n\~ao h\'a mais nada a derivar, podemos parar por aqui.} \end{array}$

_________

$\large\begin{array}{l} \textsf{(b) }\mathsf{g(x)=2x^3-\ell n\left(\dfrac{x+1}{4-x}\right)}\\\\ \textsf{Podemos enxergar uma fun\c{c}\~ao composta envolvida aqui tamb\'em.}\\ \textsf{Veja:}\\\\ \left\{\! \begin{array}{l} \mathsf{g(x)=2x^3-\ell n(v)}\\\\ \mathsf{v=\dfrac{x+1}{4-x}}\qquad\textsf{(aqui temos um quociente)} \end{array} \right. \end{array}$


$\large\begin{array}{l} \textsf{Vamos derivar usando a Regra da Cadeia:}\\\\ \mathsf{\dfrac{dg}{dx}=\dfrac{d}{dx}\left(2x^3-\ell n(v)\right)}\\\\ \mathsf{\dfrac{dg}{dx}=2\cdot 3x^{3-1}-\dfrac{d}{dx}\big(\ell n(v)\big)}\\\\ \mathsf{\dfrac{dg}{dx}=6x^2-\dfrac{d}{dv}\big(\ell n(v)\big)\cdot \dfrac{dv}{dx}}\\\\ \mathsf{\dfrac{dg}{dx}=6x^2-\dfrac{1}{v}\cdot \dfrac{dv}{dx}}\\\\ \mathsf{\dfrac{dg}{dx}=6x^2-\dfrac{1}{v}\cdot \dfrac{d}{dx}\!\left(\dfrac{x+1}{4-x}\right)} \end{array}$

$\large\begin{array}{l} \mathsf{\dfrac{dg}{dx}=6x^2-\dfrac{1}{\left(\frac{x+1}{4-x}\right)}\cdot \dfrac{\frac{d}{dx}(x+1)\cdot (4-x)-(x+1)\cdot \frac{d}{dx}(4-x)}{(4-x)^2}}\\\\ \mathsf{\dfrac{dg}{dx}=6x^2-\dfrac{4-x}{x+1}\cdot \dfrac{1\cdot (4-x)-(x+1)\cdot (-1)}{(4-x)^2}}\\\\ \mathsf{\dfrac{dg}{dx}=6x^2-\dfrac{4-x}{x+1}\cdot \dfrac{4-x+x+1}{(4-x)^2}}\\\\ \mathsf{\dfrac{dg}{dx}=6x^2-\dfrac{(4-x)\cdot 5}{(x+1)(4-x)^2}}\\\\ \boxed{\begin{array}{c}\mathsf{\dfrac{dg}{dx}=6x^2-\dfrac{5}{(x+1)(4-x)}} \end{array}} \end{array}$


$\large\textsf{D\'uvidas? Comente}\\\\\\ \textsf{Bons estudos! :-)}$