How do you find the first and second derivative of ln(x^4+5x^2)^(3/2) ?

Jun 2, 2018

$f ' \left(x\right) = \frac{3}{2} \cdot \frac{10 \cdot x + 4 {x}^{3}}{5 {x}^{2} + {x}^{4}}$
$f ' ' \left(x\right) = - \frac{3 \left(25 + 5 {x}^{2} + 2 {x}^{4}\right)}{{x}^{2} \cdot {\left(5 + {x}^{2}\right)}^{2}}$

Explanation:

By the chain and power rule we get

$f ' \left(x\right) = \frac{1}{{x}^{4} + 5 {x}^{2}} ^ \left(\frac{3}{2}\right) \cdot \frac{3}{2} \cdot {\left({x}^{4} + 5 {x}^{2}\right)}^{\frac{1}{2}} \cdot \left(4 {x}^{3} + 10 x\right)$
which can be simplified to

$f ' \left(x\right) = \frac{3 \left(10 x + 4 {x}^{3}\right)}{2 \cdot \left(5 {x}^{2} + {x}^{4}\right)}$
By the quotient rule we get

f''(x)=3/2*((10+12x^2)(5x^2+x^4)-(10x+4x^3)*(10x+4x^3))/(5x^2+x^4)^2)
which can be simplified to

$f ' ' \left(x\right) = - \frac{3 \left(25 + 5 {x}^{2} + 2 {x}^{4}\right)}{{x}^{2} \cdot {\left(5 + {x}^{2}\right)}^{2}}$

Jun 2, 2018

$f ' \left(x\right) = \frac{3 \cdot \left(10 x + 4 {x}^{3}\right)}{2 \cdot \left(5 {x}^{2} + {x}^{4}\right)}$
$f ' ' \left(x\right) = - \frac{3 \cdot 25 + 5 {x}^{2} + 2 {x}^{4}}{{x}^{2} \cdot {\left(5 + {x}^{2}\right)}^{2}}$

Explanation:

By the power and chain rule we get

$f ' \left(x\right) = \frac{1}{{x}^{4} + 5 {x}^{2}} ^ \left(\frac{3}{2}\right) \cdot \left(\frac{3}{2}\right) {\left({x}^{4} + 5 {x}^{2}\right)}^{\frac{1}{2}} \cdot \left(4 {x}^{3} + 10 x\right)$

which simplifies to

$f ' \left(x\right) = \frac{3 \left(10 x + 4 {x}^{3}\right)}{2 \left(5 {x}^{2} + {x}^{4}\right)}$
By the quotient rule we get
$f ' ' \left(x\right) = \frac{3}{2} \cdot \frac{\left(10 + 12 {x}^{2}\right) \cdot \left(5 {x}^{2} + {x}^{4}\right) - \left(10 x + 4 {x}^{3}\right) \cdot \left(10 x + 4 {x}^{3}\right)}{5 {x}^{2} + {x}^{4}} ^ 2$