# How do you find the second derivative of f(x) = e^x/x^2?

Sep 28, 2016

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = \frac{{e}^{x} \left({x}^{2} - 4 x + 6\right)}{x} ^ 4$

#### Explanation:

Let us first find the first derivative of $f \left(x\right) = {e}^{x} / {x}^{2}$ using quotient rule

that if $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$

then $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\frac{\mathrm{dg}}{\mathrm{dx}} \times h \left(x\right) - \frac{\mathrm{dh}}{\mathrm{dx}} \times g \left(x\right)}{h \left(x\right)} ^ 2$ .

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{{x}^{2} \times {e}^{x} - {e}^{x} \times 2 x}{{x}^{2}} ^ 2 = \frac{{x}^{2} {e}^{x} - 2 x {e}^{x}}{x} ^ 4 = \frac{x {e}^{x} \left(x - 2\right)}{x} ^ 4 = \frac{{e}^{x} \left(x - 2\right)}{x} ^ 3$

and second derivative $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = \frac{d}{\mathrm{dx}} \left(\frac{\mathrm{df}}{\mathrm{dx}}\right) = \frac{d}{\mathrm{dx}} \frac{{e}^{x} \left(x - 2\right)}{x} ^ 3$

= $\frac{{x}^{3} \times \left({e}^{x} \left(x - 2\right) + {e}^{x} \times 1\right) - {e}^{x} \left(x - 2\right) \times 3 {x}^{2}}{x} ^ 6$

= $\frac{{e}^{x} {x}^{3} \left(x - 2\right) + {e}^{x} {x}^{3} - 3 {x}^{2} {e}^{x} \left(x - 2\right)}{x} ^ 6$

= $\frac{{e}^{x} {x}^{2} \left(x \left(x - 2\right) + x - 3 \left(x - 2\right)\right)}{x} ^ 6$

= $\frac{{e}^{x} \left({x}^{2} - 2 x + x - 3 x + 6\right)}{x} ^ 4$

= $\frac{{e}^{x} \left({x}^{2} - 4 x + 6\right)}{x} ^ 4$