# How do you use the quotient rule to find the derivative of y=x/(3+e^x) ?

Jul 30, 2014

$y ' = \frac{1}{3 + {e}^{x}} - x {e}^{x} / \left({\left(3 + {e}^{x}\right)}^{2}\right)$

Explanation,

Using Quotient Rule,

$y = f \frac{x}{g} \left(x\right)$, then y'=(f'(x)g(x)−f(x)g'(x))/(g(x))^2

Similarly following for $y = \frac{x}{3 + {e}^{x}}$

differentiating both side with respect to $x$,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\frac{x}{3 + {e}^{x}}\right)$

y'=((3+e^x)(x)'-x(3+e^x)')/((3+e^x)^2

y'=((3+e^x)-x(e^x))/((3+e^x)^2

y'=((3+e^x)-x(e^x))/((3+e^x)^2

$y ' = \frac{1}{3 + {e}^{x}} - x {e}^{x} / \left({\left(3 + {e}^{x}\right)}^{2}\right)$