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

Question

6913 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-12T09:09:41

The answer is #=(4x^3)/(1+x^4)#

The derivative of #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)#