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

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Answer 1 · 2018-06-02T20:28:16

$f'(x)=3/2*(10*x+4x^3)/(5x^2+x^4)$
$f''(x)=-(3(25+5x^2+2x^4))/(x^2*(5+x^2)^2)$

By the chain and power rule we get

$f'(x)=1/(x^4+5x^2)^(3/2)*3/2*(x^4+5x^2)^(1/2)*(4x^3+10x)$
which can be simplified to

$f'(x)=(3(10x+4x^3))/(2*(5x^2+x^4))$
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

$f''(x)=-(3(25+5x^2+2x^4))/(x^2*(5+x^2)^2)$

Answer 2 · 2018-06-02T20:37:47

$f'(x)=(3*(10x+4x^3))/(2*(5x^2+x^4))$
$f''(x)=-(3*25+5x^2+2x^4)/(x^2*(5+x^2)^2)$

By the power and chain rule we get

$f'(x)=1/(x^4+5x^2)^(3/2)*(3/2)(x^4+5x^2)^(1/2)*(4x^3+10x)$

which simplifies to

$f'(x)=(3(10x+4x^3))/(2(5x^2+x^4))$
By the quotient rule we get
$f''(x)=3/2*((10+12x^2)*(5x^2+x^4)-(10x+4x^3)*(10x+4x^3))/(5x^2+x^4)^2$

How do you find the first and second derivative of ln(x^4+5x^2)^(3/2) ln(x4+5x2)32ln(x^4+5x^2)^(3/2) ?