# How do you differentiate f(x) =x/(e^(3-x)+x^3) using the quotient rule?

Jan 24, 2018

Using the quotient rule, given a function $f \left(x\right) = g \frac{x}{h \left(x\right)}$, then the derivative f'(x)" will be

$f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{{h}^{2} \left(x\right)}$. (1)

In this case, $g \left(x\right) = x$ and $h \left(x\right) = {e}^{3 - x} + {x}^{3}$. Then:

$g ' \left(x\right)$ = 1;

h'(x) = -e^(3-x) + 3x^2;#

${\left[h \left(x\right)\right]}^{2} = {\left({e}^{3 - x} + {x}^{3}\right)}^{2}$.

Putting all these results into Equation (1):

$f ' \left(x\right) = \frac{{e}^{3 - x} + {x}^{3} - x \left(- {e}^{3 - x} + 3 {x}^{2}\right)}{{e}^{3 - x} + {x}^{3}} ^ 2$;

$f ' \left(x\right) = \frac{{e}^{3 - x} + {x}^{3} + x {e}^{3 - x} - 3 {x}^{3}}{{e}^{3 - x} + {x}^{3}} ^ 2$;

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