# Using the limit definition, how do you differentiate #f(x) = x^(1/2) #?

The crucial step is

Here is the algebra of the crucial step of "rationalizing" the numerator:

# = ((x+h)-x)//(h((x+h)^(1/2) + x^(1/2)))

Using this algebra, we get:

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To differentiate ( f(x) = x^{\frac{1}{2}} ) using the limit definition, we first write the difference quotient:

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

Substitute ( f(x) = x^{\frac{1}{2}} ) into the difference quotient:

[ f'(x) = \lim_{h \to 0} \frac{(x+h)^{\frac{1}{2}} - x^{\frac{1}{2}}}{h} ]

Next, we use the binomial expansion to simplify ( (x+h)^{\frac{1}{2}} ):

[ (x+h)^{\frac{1}{2}} = x^{\frac{1}{2}} + \frac{1}{2}h x^{-\frac{1}{2}} + O(h^2) ]

Substitute the expansion into the difference quotient:

[ f'(x) = \lim_{h \to 0} \frac{x^{\frac{1}{2}} + \frac{1}{2}h x^{-\frac{1}{2}} + O(h^2) - x^{\frac{1}{2}}}{h} ]

Simplify:

[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{2}h x^{-\frac{1}{2}} + O(h^2)}{h} ]

[ f'(x) = \lim_{h \to 0} \left(\frac{\frac{1}{2}h x^{-\frac{1}{2}}}{h} + \frac{O(h^2)}{h}\right) ]

[ f'(x) = \lim_{h \to 0} \left(\frac{\frac{1}{2} x^{-\frac{1}{2}}}{1} + \frac{O(h^2)}{h}\right) ]

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

Therefore, the derivative of ( f(x) = x^{\frac{1}{2}} ) using the limit definition is ( f'(x) = \frac{1}{2} x^{-\frac{1}{2}} ).

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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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