# How do you find f'(x) using the definition of a derivative #f(x) = sqrtx + 2#?

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To find ( f'(x) ) using the definition of a derivative for the function ( f(x) = \sqrt{x} + 2 ), we first recall the definition:

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

Now, we substitute ( f(x) = \sqrt{x} + 2 ) into the definition:

[ f'(x) = \lim_{{h \to 0}} \frac{{\sqrt{x + h} + 2 - (\sqrt{x} + 2)}}{h} ]

Next, we simplify the expression:

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

To eliminate the square roots from the numerator, we multiply the expression by the conjugate of the numerator:

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

This simplifies to:

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

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

Now, we can evaluate the limit:

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

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

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