How do you find the second derivative of $f (x) = \frac{e^{x}}{x^{2}}$ $f(x) = e^x/x^2$ ?

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How do you find the second derivative of $f (x) = \frac{e^{x}}{x^{2}}$ $f(x) = e^x/x^2$ ?

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Answer 1 · 2016-09-28T07:20:19

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

Explanation:

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

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

then $\frac{d f}{d x} = \frac{\frac{d g}{d x} \times h (x) - \frac{d h}{d x} \times g (x)}{{(h (x))}^{2}}$ $(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2$ .

$\frac{d f}{d x} = \frac{x^{2} \times e^{x} - e^{x} \times 2 x}{{(x^{2})}^{2}} = \frac{x^{2} e^{x} - 2 x e^{x}}{x^{4}} = \frac{x e^{x} (x - 2)}{x^{4}} = \frac{e^{x} (x - 2)}{x^{3}}$ $(df)/(dx)=(x^2xxe^x-e^x xx2x)/(x^2)^2=(x^2e^x-2xe^x)/x^4=(xe^x(x-2))/x^4=(e^x(x-2))/x^3$

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

= $\frac{x^{3} \times (e^{x} (x - 2) + e^{x} \times 1) - e^{x} (x - 2) \times 3 x^{2}}{x^{6}}$ $(x^3xx(e^x(x-2)+e^x xx1)-e^x(x-2)xx3x^2)/x^6$

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

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

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

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

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How do you find the second derivative of $f (x) = \frac{e^{x}}{x^{2}}$ $f(x) = e^x/x^2$ ?

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