What is the second derivative of #f(x)= ln (x^2+2x)#?

Answer 1

#f''(x) =-4(x+1)/(x^2+2)^2#

Use the substitution #t = x^2+2x# Then #f(x) = lnt#
#(df)/dt = 1/t# #(dt)/dx =2x +2#
#(df)/dx = (df)/dt * (dt)/dx = (1/(x^2+2x))* (2x+2)# #f'(x) = (2(x+1))/(x^2+2x)#

We can now use the quotient rule to find the second derivative

#f''(x) = ((x^2+2x)*2 - 2(x+1)*(2x+2))/(x^2+2)^2#
#f''(x) = (2x^2 +4x -2x^2 - 8x -4)/(x^2+2)^2# #f''(x) =-4(x+1)/(x^2+2)^2#
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Answer 2

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))

  1. 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} ]

  2. 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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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.

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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