What is the second derivative of #f(x)= ln (x^2+2x)#?
We can now use the quotient rule to find the second derivative
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To find the second derivative of (f(x) = \ln(x^2 + 2x)), we first find the first derivative using the chain rule, then differentiate again:
Given: (f(x) = \ln(x^2 + 2x))

Find the first derivative, (f'(x)): [ \begin{aligned} f'(x) &= \frac{d}{dx}[\ln(x^2 + 2x)] \ &= \frac{1}{x^2 + 2x} \cdot \frac{d}{dx}(x^2 + 2x) \ &= \frac{1}{x^2 + 2x} \cdot (2x + 2) \ &= \frac{2x + 2}{x^2 + 2x} \end{aligned} ]

Find the second derivative, (f''(x)): [ \begin{aligned} f''(x) &= \frac{d}{dx}\left(\frac{2x + 2}{x^2 + 2x}\right) \ &= \frac{(x^2 + 2x)\cdot 2  (2x + 2)\cdot (2x + 1)}{(x^2 + 2x)^2} \ &= \frac{2x^2 + 4x  4x^2  4x  2}{(x^2 + 2x)^2} \ &= \frac{2x^2  8x  2}{(x^2 + 2x)^2} \ &= \frac{2(x^2 + 4x + 1)}{(x^2 + 2x)^2} \end{aligned} ]
Therefore, the second derivative of (f(x) = \ln(x^2 + 2x)) is (f''(x) = \frac{2(x^2 + 4x + 1)}{(x^2 + 2x)^2}).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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