# How do you find the derivative of #y = sqrt(x + 1)# using the limit definition?

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To find the derivative of ( y = \sqrt{x + 1} ) using the limit definition, we start with the definition of the derivative:

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

Substitute ( f(x) = \sqrt{x + 1} ) into the definition:

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

This simplifies to:

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

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

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

Now, as ( h ) approaches 0, the expression ( \sqrt{x + 1 + h} ) approaches ( \sqrt{x + 1} ), so:

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

Thus, the derivative of ( y = \sqrt{x + 1} ) with respect to ( x ) using the limit definition is:

[ \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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