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

It's going to demand chain rule, which states that

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To find the derivative of ( y = \ln((x^2) + 1) ), you can use the chain rule. The chain rule states that if ( y = f(g(x)) ), then ( y' = f'(g(x)) \cdot g'(x) ). In this case, ( f(u) = \ln(u) ) and ( g(x) = x^2 + 1 ).

So, the derivative of ( f(u) = \ln(u) ) with respect to ( u ) is ( f'(u) = \frac{1}{u} ). And the derivative of ( g(x) = x^2 + 1 ) with respect to ( x ) is ( g'(x) = 2x ).

Now, applying the chain rule:

[ \frac{dy}{dx} = \frac{1}{x^2 + 1} \cdot 2x ]

Simplifying this, you get:

[ \frac{dy}{dx} = \frac{2x}{x^2 + 1} ]

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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.

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.

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.

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