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

If we rewrite using laws of logarithms, we get:

By the chain rule, we get:

Now by the quotient rule, we get:

Hopefully this helps!

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To find the second derivative of ( f(x) = \ln\sqrt{x + e^x} ), you first need to find the first derivative and then differentiate it again.

Given: [ f(x) = \ln\sqrt{x + e^x} ]

First, let's find the first derivative: [ f'(x) = \frac{d}{dx}\left(\ln\sqrt{x + e^x}\right) ]

Using the chain rule and the derivative of natural logarithm: [ f'(x) = \frac{1}{\sqrt{x + e^x}} \cdot \frac{1}{2\sqrt{x + e^x}} \cdot \frac{d}{dx}(x + e^x) ]

[ = \frac{1}{2(x + e^x)} \cdot \frac{1}{\sqrt{x + e^x}} \cdot (1 + e^x) ]

Now, we'll find the second derivative: [ f''(x) = \frac{d}{dx}\left(\frac{1}{2(x + e^x)} \cdot \frac{1}{\sqrt{x + e^x}} \cdot (1 + e^x)\right) ]

[ = \frac{d}{dx}\left(\frac{1 + e^x}{2(x + e^x) \sqrt{x + e^x}}\right) ]

Using the quotient rule: [ = \frac{(2(x + e^x) \sqrt{x + e^x})\cdot 0 - (1 + e^x)\cdot \left(\frac{2\sqrt{x + e^x} + (1 + e^x)\frac{1}{2\sqrt{x + e^x}}}{(2(x + e^x) \sqrt{x + e^x})^2}\right)}{(2(x + e^x) \sqrt{x + e^x})^2} ]

[ = -\frac{(1 + e^x)(2\sqrt{x + e^x} + (1 + e^x)\frac{1}{2\sqrt{x + e^x}})}{4(x + e^x)(x + e^x)} ]

[ = -\frac{(1 + e^x)\left(2\sqrt{x + e^x} + \frac{1 + e^x}{2\sqrt{x + e^x}}\right)}{4(x + e^x)(x + e^x)} ]

[ = -\frac{(1 + e^x)\left(4\sqrt{x + e^x} + 1 + e^x\right)}{8(x + e^x)(x + e^x)} ]

Thus, the second derivative of ( f(x) = \ln\sqrt{x + e^x} ) is: [ f''(x) = -\frac{(1 + e^x)\left(4\sqrt{x + e^x} + 1 + e^x\right)}{8(x + e^x)(x + e^x)} ]

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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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