# How do you find the derivative using limits of #f(x)=sqrt(x+1)#?

There are a few different versions of the limit definition of a derivative. For this answer, I'll opt for the following definition:

Thus, we continue:

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To find the derivative of ( f(x) = \sqrt{x + 1} ), you can use the definition of the derivative:

[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]

First, substitute ( f(x) = \sqrt{x + 1} ) into the formula:

[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h + 1} - \sqrt{x + 1}}{h} ]

To simplify, multiply the numerator and denominator by the conjugate of the numerator:

[ f'(x) = \lim_{h \to 0} \frac{(\sqrt{x + h + 1} - \sqrt{x + 1})(\sqrt{x + h + 1} + \sqrt{x + 1})}{h(\sqrt{x + h + 1} + \sqrt{x + 1})} ]

[ = \lim_{h \to 0} \frac{(x + h + 1) - (x + 1)}{h(\sqrt{x + h + 1} + \sqrt{x + 1})} ]

[ = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h + 1} + \sqrt{x + 1})} ]

[ = \lim_{h \to 0} \frac{1}{\sqrt{x + h + 1} + \sqrt{x + 1}} ]

Now, substitute ( h = 0 ) into the expression:

[ f'(x) = \frac{1}{\sqrt{x + 1} + \sqrt{x + 1}} ]

[ f'(x) = \frac{1}{2\sqrt{x + 1}} ]

So, the derivative of ( f(x) = \sqrt{x + 1} ) is ( f'(x) = \frac{1}{2\sqrt{x + 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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