How do you find the derivative of $ln (1 + x^{4})$ $ln(1+x^4)$ ?

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Answer 1 · 2017-02-12T09:09:41

The answer is $= \frac{4 x^{3}}{1 + x^{4}}$ $=(4x^3)/(1+x^4)$

The derivative of $ln (u (x))$ $ln(u(x))$ is

$=(u'(x))/(u(x))$

Here,

$u(x)=1+x^4$

$u'(x)=4x^3$

So,

$(ln(1+x^4))'=1/(1+x^4)*4x^3$

$=(4x^3)/(1+x^4)$

Answer 2 · 2017-02-12T09:09:54

You use the natural log equation for derivatives

if $y = lnf(x)$
then $y' = 1/f(x) * f'(x)$

So, in this case,

$y = ln(1+x^4)$

$y' = 1/(1+x^4)*(1+x^4)'$

$y' = (4x^3)/(1+x^4)$

How do you find the derivative of ln(1+x^4)ln(1+x4)ln(1+x^4)?