# What is the second derivative of f(x)= sqrt(5+x^6)/x?

Mar 9, 2017

$\frac{2 {x}^{12} + 65 {x}^{6} + 50}{{x}^{3} \left(5 + {x}^{6}\right) \left(\sqrt{5 + {x}^{6}}\right)}$

#### Explanation:

The second derivative is the derivative of the first derivative:

$f ' \left(x\right) = \frac{\frac{1}{\cancel{2} \sqrt{5 + {x}^{6}}} \cdot {\cancel{6}}^{3} {x}^{5} \cdot x - \sqrt{5 + {x}^{6}} \cdot 1}{x} ^ 2$

$= \frac{\frac{3 {x}^{6}}{\sqrt{5 + {x}^{6}}} - \sqrt{5 + {x}^{6}}}{x} ^ 2$

$= \frac{\frac{3 {x}^{6} - 5 - {x}^{6}}{\sqrt{5 + {x}^{6}}}}{x} ^ 2$

$= \frac{2 {x}^{6} - 5}{{x}^{2} \sqrt{5 + {x}^{6}}}$

f''(x)=(12x^5*x^2sqrt(5+x^6)-(2x^6-5)*(2xsqrt(5+x^6)+x^2/(cancel2sqrt(5+x^6))*cancel6^3x^5))/(x^4(5+x^6)

=(12x^7sqrt(5+x^6)-(2x^6-5)*(2xsqrt(5+x^6)+(3x^7)/(sqrt(5+x^6))))/(x^4(5+x^6)

$= \frac{12 {x}^{7} \sqrt{5 + {x}^{6}} - \left(2 {x}^{6} - 5\right) \cdot \frac{2 x \left(5 + {x}^{6}\right) + 3 {x}^{7}}{\sqrt{5 + {x}^{6}}}}{{x}^{4} \left(5 + {x}^{6}\right)}$

$= \frac{12 {x}^{7} \sqrt{5 + {x}^{6}} - \left(2 {x}^{6} - 5\right) \cdot \frac{10 x + 2 {x}^{7} + 3 {x}^{7}}{\sqrt{5 + {x}^{6}}}}{{x}^{4} \left(5 + {x}^{6}\right)}$

$= \frac{12 {x}^{7} \sqrt{5 + {x}^{6}} - \left(2 {x}^{6} - 5\right) \cdot \frac{10 x + 5 {x}^{7}}{\sqrt{5 + {x}^{6}}}}{{x}^{4} \left(5 + {x}^{6}\right)}$

$= \frac{\frac{12 {x}^{7} \left(5 + {x}^{6}\right) - \left(2 {x}^{6} - 5\right) \cdot \left(10 x + 5 {x}^{7}\right)}{\sqrt{5 + {x}^{6}}}}{{x}^{4} \left(5 + {x}^{6}\right)}$

$= \frac{12 {x}^{7} \left(5 + {x}^{6}\right) - \left(2 {x}^{6} - 5\right) \cdot \left(10 x + 5 {x}^{7}\right)}{{x}^{4} \left(5 + {x}^{6}\right) \left(\sqrt{5 + {x}^{6}}\right)}$

$= \frac{60 {x}^{7} + 12 {x}^{13} - 20 {x}^{7} - 10 {x}^{13} + 50 x + 25 {x}^{7}}{{x}^{4} \left(5 + {x}^{6}\right) \left(\sqrt{5 + {x}^{6}}\right)}$

$= \frac{2 {x}^{13} + 65 {x}^{7} + 50 x}{{x}^{4} \left(5 + {x}^{6}\right) \left(\sqrt{5 + {x}^{6}}\right)}$

$= \frac{2 {x}^{12} + 65 {x}^{6} + 50}{{x}^{3} \left(5 + {x}^{6}\right) \left(\sqrt{5 + {x}^{6}}\right)}$