Question

Derivada de
$y=e ^\frac-{1}{2}^(^x^-^1^)^2$

Obs: Esse 2 do (x-1) é uma potência, não deu para colocar. Fica

$(x-1)^2$

Answer 1

Temos que:

$Y = e^-^1^/^2^(^x^-^1^)^2$

$\\ Y' = [ e^-^1^/^2^(^x^-^1^)^2]'*[- \frac{1}{2} *(x-1)^2]'*(x-1)' \\ \\ Y' = e^-^1^/^2^(^x^-^1^)^2*[-2* \frac{1}{2} *(x-1)^2^-^1]*(1-0) \\ \\ Y' = e^-^1^/^2^(^x^-^1^)^2*[-(x-1)^1]*1 \\ \\ Y' = e^-^1^/^2^(^x^-^1^)^2-(x-1) \\ \\ Y' = -(x-1)e^-^1^/^2^(^x^-^1^)^2 \\ \\ Y' = \frac{-(x-1)}{e^1^/^2^(^x^-^1^)^2} \\ \\ Y' = - \frac{x-1}{e^1^/^2^(^x^-^1^)^2}$

Answer 2

$\sf \displaystyle y=e\:\frac{-1}{2}\left(x-1\right)^{^2}\\\\\\\frac{d}{dx}\left(e\frac{-1}{2}\left(x-1\right)^2\right)\\\\\\=e\frac{-1}{2}\frac{d}{dx}\left(\left(x-1\right)^2\right)\\\\\\=2\left(x-1\right)\frac{d}{dx}\left(x-1\right)\\\\\\=e\frac{-1}{2}\cdot \:2\left(x-1\right)\cdot \:1\\\\\\\to \boxed{\sf =-e\left(x-1\right)}$