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

Answer 1

Elijah Abram

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)#

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Answer 2

Christopher Allers

To find the derivative of ( \ln{(x^2 + 1)}^{\frac{1}{2}} ), you can use the chain rule. The derivative is:

[ \frac{d}{dx} \left( \ln{(x^2 + 1)}^{\frac{1}{2}} \right) = \frac{1}{2\sqrt{\ln{(x^2 + 1)}}} \cdot \frac{d}{dx} \ln{(x^2 + 1)} ]

Now, applying the chain rule for ( \ln{(x^2 + 1)} ):

[ \frac{d}{dx} \ln{(x^2 + 1)} = \frac{1}{x^2 + 1} \cdot \frac{d}{dx}(x^2 + 1) = \frac{2x}{x^2 + 1} ]

Substitute this back into the original expression:

[ \frac{d}{dx} \left( \ln{(x^2 + 1)}^{\frac{1}{2}} \right) = \frac{1}{2\sqrt{\ln{(x^2 + 1)}}} \cdot \frac{2x}{x^2 + 1} ]

Simplify to get the final derivative:

[ \frac{x}{(x^2 + 1)\sqrt{\ln{(x^2 + 1)}}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.