Question

Para os mestres do Calculo:Sejam f e g funcões diferenciaveis e tais que f(x) =g[(e^x²) - x)/x² + 1. Sabendo que  g'(1) =15,determine f '(0).

Answer 1

$\large\begin{array}{l} \mathsf{f(x)=g\bigg(\dfrac{e^{x^2}-x}{x^2+1}\bigg)}\\\\ \mathsf{f(x)=g\big(h(x)\big)}\\\\\\ \textsf{sendo }\mathsf{h(x)=\dfrac{e^{x^2}-x}{x^2+1}.} \end{array}$


$\large\begin{array}{l} \textsf{Computemos h(0):}\\\\ \mathsf{h(0)=\dfrac{e^{0^2}-0}{0^2+1}}\\\\ \mathsf{h(0)=\dfrac{e^0-0}{0+1}}\\\\ \mathsf{h(0)=\dfrac{1-0}{0+1}}\\\\ \mathsf{h(0)=\dfrac{1}{1}}\\\\ \mathsf{h(0)=1} \end{array}$


$\large\begin{array}{l} \textsf{Vamos derivar f com rela\c{c}\~ao a x, usando a Regra da Cadeia:}\\\\ \mathsf{f'(x)=\big[g\big(h(x)\big)\big]'}\\\\ \mathsf{f'(x)=g'\big(h(x)\big)\cdot h'(x)}\\\\ \mathsf{f'(x)=g'\big(h(x)\big)\cdot\bigg(\dfrac{e^{x^2}-x}{x^2+1}\bigg)'}\\\\ \mathsf{f'(x)=g'\big(h(x)\big)\cdot\dfrac{(e^{x^2}-x)'\cdot (x^2+1)-(e^{x^2}-x)\cdot (x^2+1)'}{(x^2+1)^2}}\\\\ \mathsf{f'(x)=g'\big(h(x)\big)\cdot\dfrac{\big[(e^{x^2})'-1\big]\cdot (x^2+1)-(e^{x^2}-x)\cdot (2x+0)}{(x^2+1)^2}}\\\\ \mathsf{f'(x)=g'\big(h(x)\big)\cdot\dfrac{\big[e^{x^2}\cdot (x^2)'-1\big]\cdot (x^2+1)-(e^{x^2}-x)\cdot 2x}{(x^2+1)^2}}\\\\ \mathsf{f'(x)=g'\big(h(x)\big)\cdot\dfrac{\big[e^{x^2}\cdot 2x-1\big]\cdot (x^2+1)-(e^{x^2}-x)\cdot 2x}{(x^2+1)^2}}\\\\ \mathsf{f'(x)=g'\big(h(x)\big)\cdot\dfrac{(2x\,e^{x^2}-1)\cdot (x^2+1)-(e^{x^2}-x)\cdot 2x}{(x^2+1)^2}}\\\\ \end{array}$


$\large\begin{array}{l} \textsf{Para }\mathsf{x=0,}\textsf{ ficamos com}\\\\ \mathsf{f'(0)=g'\big(h(0)\big)\cdot\dfrac{(2\cdot 0\cdot e^{0^2}-1)\cdot (0^2+1)-(e^{0^2}-0)\cdot 2\cdot 0}{(0^2+1)^2}}\\\\ \mathsf{f'(0)=g'(1)\cdot\dfrac{(0-1)\cdot (0+1)-0}{(0+1)^2}}\\\\ \mathsf{f'(0)=15\cdot\dfrac{(-1)\cdot 1}{1^2}}\\\\ \mathsf{f'(0)=15\cdot\dfrac{(-1)}{1}}\\\\ \boxed{\begin{array}{c}\mathsf{f'(0)=-15} \end{array}} \end{array}$


$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$