How do you find the derivative of #ln(x^2+1)#?

Answer 1

#frac{d}{dx}(y)=frac{2x}{x^2+1}#

Given equation is #y=ln(x^2+1)#

One fundamental equation of derivative calculus is #frac{d}{dx}(y)=1/x# when #y=lnx#

Another important aspect of derivative calculus is the chain rule, that is if #y=f(t)# and #t=g(x)#, then #dy/dx=dy/dt*dt/dx#

Taking #t=x^2+1#, from chain rule

#dy/dx=\frac{d}{dt}(lnt)*dt/dx# #\implies dy/dx=1/t*fracddx(x^2+1)#

Substitute for whatever #t# remains in the equation as #x^2+1# and you'll get the answer.

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Answer 2

Aurora Alvarado

To find the derivative of ln(x^2+1), you can use the chain rule. The derivative of ln(u) with respect to x is du/dx divided by u. Here, u = x^2 + 1. So, the derivative of ln(x^2+1) with respect to x is (2x)/(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.