What is the derivative of #ln(x^2+1)^(1/2)#?

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Answer 1 · 2015-05-14T16:38:41

Here, we must resort to a three-part chain rule.

First, we must rename the whole problem.

#u=x^2+1#, #w=u^(1/2)# and #z=ln(w)#

The derivatives of these three are:

#u'=2x#, #w'=1/(2*u^(1/2)# and #z'=(w')/w#

Now, multiplying all the derivatives...

#(dy)/(dx)=z'*w'*u'#

#(dy)/(dx) = (w')/(w)*1/(2u^(1/2))*2x#

#(dy)/(dx) = (1/(2u))*(1/(2u^(1/2)))*2x#

Finally:

#(dy)/(dx) = x/(2(x^2-1)^(1/2)#