Question

Bom dia ,derivada y= [(x+1])^3\ x^3\2 ?

Answer 1

$y = \frac{(x+1)^3}{x^3^/^2}$

$\\ y' = \frac{[(x+1)^3]'*x^3^/^2-(x+1)^3*(x^3^/^2)'}{(x^3^/^2)^2} \\ \\ y' = \frac{3(x+1)^2*x^3^/^2-(x+1)^3* \frac{3}{2}*x^1^/^2 }{x^3} \\ \\ y' = \frac{3x^3^/^2(x+1)^2- \frac{3}{2}x^1^/^2(x+1)^3 }{x^3} \\ \\ y' = \frac{3*x^1*x^1^/^2(x+1)^2- \frac{3}{2}x^1^/^2(x+1)^3 }{x^3} \\ \\ y' = \frac{3x^1^/^2[x(x+1)^2- \frac{1}{2}(x+1)^3] }{x^3} \\ \\ y'= 3x^-^5^/^2 (x+1)^2[x - \frac{1}{2} (x+1)] \\ \\ y' = \frac{3(x+1)^2[ \frac{2x-1}{2} *(x+1)]}{x^5^/^2}$

$\\ y' = \frac{3}{2} * \frac{(x+1)^2(2x-1)*(x+1)}{x^5^/^2} \\ \\ y' = \frac{3(x+1)^3(2x-1)}{2x^5^/^2}$