What is the second derivative of #f(x)= ln (x^3+e^x)#?
We first use the log rule to find
We now use the quotient rule to find
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To find the second derivative of ( f(x) = \ln(x^3 + e^x) ), follow these steps:
 Find the first derivative of ( f(x) ) using the chain rule and the derivative of the natural logarithm function.
 Once you have the first derivative, differentiate it again to find the second derivative.
Here's the process:

First derivative: [ f'(x) = \frac{d}{dx}(\ln(x^3 + e^x)) ] [ f'(x) = \frac{1}{x^3 + e^x} \cdot \frac{d}{dx}(x^3 + e^x) ] [ f'(x) = \frac{1}{x^3 + e^x} \cdot (3x^2 + e^x) ]

Second derivative: [ f''(x) = \frac{d}{dx}\left(\frac{1}{x^3 + e^x} \cdot (3x^2 + e^x)\right) ] [ f''(x) = \frac{d}{dx}\left(\frac{3x^2 + e^x}{x^3 + e^x}\right) ] [ f''(x) = \frac{(6x + e^x)(x^3 + e^x)  (3x^2 + e^x)(3x^2)}{(x^3 + e^x)^2} ] [ f''(x) = \frac{(6x + e^x)(x^3 + e^x)  9x^4  3x^2e^x}{(x^3 + e^x)^2} ]
So, the second derivative of ( f(x) = \ln(x^3 + e^x) ) is given by: [ f''(x) = \frac{(6x + e^x)(x^3 + e^x)  9x^4  3x^2e^x}{(x^3 + e^x)^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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