How do I use the limit definition of derivative to find #f'(x)# for #f(x)=1/sqrt(x)# ?

Answer 1
#f(x)=1/sqrt(x)=1/x^(1/2)=x^(-1/2)#
Note that a square root is equivalent to raising an expression to the #1/2# power.
#f'(x)=lim_(h->0) (f(x+h)-f(x))/h#
#f(x+h)=1/sqrt(x+h)#
#f'(x)=lim_(h->0) (1/sqrt(x+h)-1/sqrt(x))/h#

Find the common denominator

#f'(x)=lim_(h->0) (1/sqrt(x+h)*sqrt(x)/sqrt(x)-1/sqrt(x)*sqrt(x+h)/sqrt(x+h))/h#
#f'(x)=lim_(h->0) (sqrt(x)/(sqrt(x)sqrt(x+h))-sqrt(x+h)/(sqrt(x)sqrt(x+h)))/h#

Consolidate the numerator of the complex fraction.

#f'(x)=lim_(h->0) ((sqrt(x)-sqrt(x+h))/(sqrt(x)sqrt(x+h)))/h#

Dividing fractions is equivalent to multiplying by the reciprocal

#f'(x)=lim_(h->0) 1/h*((sqrt(x)-sqrt(x+h))/(sqrt(x)sqrt(x+h)))#

Rationalize the numerator

#f'(x)=lim_(h->0) 1/h*((sqrt(x)-sqrt(x+h))/(sqrt(x)sqrt(x+h)))*(sqrt(x)+sqrt(x+h))/(sqrt(x)+sqrt(x+h)#

Simplify. Remember difference of perfect squares.

#f'(x)=lim_(h->0) 1/h*((x-(x+h))/(sqrt(x)sqrt(x+h)(sqrt(x)+sqrt(x+h))))#

Distribute the negative in the numerator

#f'(x)=lim_(h->0) 1/h*(x-x-h)/(sqrt(x)sqrt(x+h)(sqrt(x)+sqrt(x+h)))#
The #x's# resolve to zero.
#f'(x)=lim_(h->0) 1/h*(-h)/(sqrt(x)sqrt(x+h)(sqrt(x)+sqrt(x+h)))#
The #h's# can be cancelled.
#f'(x)=lim_(h->0) (-1)/(sqrt(x)sqrt(x+h)(sqrt(x)+sqrt(x+h)))#
Now we can substitute in 0 for #h#.
#(-1)/(sqrt(x)sqrt(x+0)(sqrt(x)+sqrt(x+0)))#
#(-1)/(sqrt(x)sqrt(x)(sqrt(x)+sqrt(x)))#

Manipulate the exponents

#(-1)/(x(2sqrt(x)))=(-1)/(x^1*2*x^(1/2))=(-1)/(x^(2/2)*2*x^(1/2))=(-1)/(2x^(3/2))#
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Answer 2

To use the limit definition of the derivative to find ( f'(x) ) for ( f(x) = \frac{1}{\sqrt{x}} ), we start with the definition of the derivative:

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

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

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

To simplify, rationalize the numerator:

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

Now, factor out ( \sqrt{x} ) from the numerator:

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

Apply the conjugate rule to simplify the numerator:

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

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

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

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

Now, as ( h ) approaches 0, the denominator remains finite, so we can evaluate the limit:

[ f'(x) = \frac{-1}{2x^{3/2}} ]

Therefore, the derivative of ( f(x) = \frac{1}{\sqrt{x}} ) is ( f'(x) = \frac{-1}{2x^{3/2}} ).

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Answer from HIX Tutor

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.

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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