How do you differentiate f(x)=(x^2+e^x)/(x-1)^2 using the quotient rule?

Nov 17, 2015

$\frac{d}{\mathrm{dx}} \frac{{x}^{2} + {e}^{x}}{x - 1} ^ 2 = \frac{{\left(x - 1\right)}^{2} \cdot \left(2 x + {e}^{x}\right) - 2 \left(x - 1\right) \cdot \left({x}^{2} + {e}^{x}\right)}{{\left(x - 1\right)}^{2}}$

Explanation:

The quotient rule states that
$\frac{d}{\mathrm{dx}} f \frac{x}{g \left(x\right)} = \frac{g \left(x\right) \cdot f ' \left(x\right) - f \left(x\right) \cdot g ' \left(x\right)}{g \left(x\right)} ^ 2$

So in this case,

$\frac{d}{\mathrm{dx}} \frac{{x}^{2} + {e}^{x}}{x - 1} ^ 2 = \frac{{\left(x - 1\right)}^{2} \cdot \left(2 x + {e}^{x}\right) - 2 \left(x - 1\right) \cdot \left({x}^{2} + {e}^{x}\right)}{{\left(x - 1\right)}^{2}}$