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

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Answer 1 · 2018-01-24T18:31:18

Using the quotient rule, given a function $f(x) = g(x)/(h(x))$ , then the derivative #f'(x)" will be

$f'(x) = (g'(x)h(x) - g(x)h'(x))/(h^2(x))$ . (1)

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

$g'(x)$ = 1;

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

$[h(x)]^2 = (e^(3-x) + x^3)^2$ .

Putting all these results into Equation (1):

$f'(x) = (e^(3-x) + x^3 - x(-e^(3-x) + 3x^2))/(e^(3-x) + x^3)^2$ ;

$f'(x) = (e^(3-x) + x^3 + xe^(3-x) - 3x^3)/(e^(3-x) + x^3)^2$ ;

$f'(x) = (e^(3-x)(x+1) - 2x^3)/(e^(3-x) + x^3)^2$ .

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