Question

derive a função

y= (x+1)²/ (1-x)





a resposta é y'= 4(x+1) / (1-x)³ ( não conseguir formular o calculo para dar esse resultado)

Answer 1

regras de derivação que vc vai usar:

$\bmatrix \left( \frac{U}{V} \right)'= \frac{U'*V-U*V'}{V^2}\\\\ (U^N)' = N*U^{N-1}*U' \end$


$y= \frac{(x+1)^2}{(1-x)} \\\\ y'= \frac{((x+1)^2)'*(1-x) - (x+1)^2*(1-x)'}{(1-x)^2} \\\\ y'= \frac{(2(x+1)*(1+0))*(1-x)-(x+1)^2*(0-1)}{(1-x)^2} \\\\ y'= \frac{2(x+1)(1-x)+(x+1)^2}{(1-x)^2} \\\\ y'= \frac{(x+1)*[2(1-x)+(x+1)]}{(1-x)^2} \\\\ y'= \frac{(x+1)*[2-2x+x+1]}{(1-x)^2} \\\\ y'= \frac{(x+1)*[3-x]}{(1-x)^2} \\\\\boxed{\boxed{y'= \frac{(x+1)(3-x)}{(1-x)^2} }}$

Answer 2

Olá


$\displaystyle \mathsf{y= \frac{(x+1)^2}{(1-x)^2} }\\\\\\\text{Para derivar essa funcao, temos que aplicar a regra do quociente, dada}\\\text{por:}\\\\\mathsf{(\frac{f}{g})' = \frac{f'\cdot g -f\cdot g'}{g^2} }\\\\\\\text{Aplicando essa regra}\\\\\\ \mathsf{y'= \frac{(2\cdot(x+1)^{2-1}\cdot1\cdot(1-x)^2)~-~((x+1)^2\cdot 2\cdot(1-x)^{2-1}\cdot (-1))}{((1-x)^{2})^2} }\\\\\\\mathsf{y'= \frac{2\cdot(x+1)\cdot (1-x)^2~-~(-2\cdot (x+1)^2\cdot (1-x))}{(1-x)^4} }$

$\displaystyle \mathsf{y'= \frac{2\cdot(x+1)\cdot (1-x)^2+2\cdot (x+1)^2\cdot (1-x)}{(1-x)^4} }\\\\\\\text{Poe o termo (1-x) em evidencia}\\\\\\\mathsf{y'= \frac{(1-x)(2\cdot(x+1)\cdot (1-x)+2\cdot (x+1)^2)}{(1-x)^4} }\\\\\\\text{Simplifica com o denominador}\\\\\\\mathsf{y'= \frac{(\diagup\!\!\!\!1-\diagup\!\!\!\!x)(2\cdot(x+1)\cdot (1-x)+2\cdot (x+1)^2)}{(1-x)^{\diagup\!\!\!\!4}} }\\\\\\\mathsf{y'= \frac{2\cdot(x+1)\cdot (1-x)+2\cdot (x+1)^2}{(1-x)^3} }\\\\\\\text{Expande os quadrados}$

$\displaystyle \mathsf{y'= \frac{(2\cdot(x+1)\cdot (1-x))~+~(2\cdot (x^2+2x+1))}{(1-x)^3} }\\\\\\\mathsf{y'= \frac{(2-2x^2)~+~(2x^2+4x+2)}{(1-x)^3} }\\\\\\\text{Simplifica e agrupa os termos em comuns}\\\\\\\mathsf{y'= \frac{2-\diagup\!\!\!\!\!\!2x^2~+~\diagup\!\!\!\!\!\!2x^2+4x+2}{(1-x)^3} }\\\\\\\mathsf{y'= \frac{4x+4}{(1-x)^3} }\\\\\\\text{Poe o 4 em evidencia}\\\\\\\boxed{\mathsf{y'= \frac{4(x+1)}{(1-x)^3} }}$