Find the derivative using first principles? : #sqrt(x-5) #

Answer 1

# d/dx sqrt(x-5) = 1/(2sqrt(x-5)) #

We seek:

# d/dx sqrt(x-5) #

By First Principles, using the limit definition:

# L = lim_(h rarr 0) (sqrt((x+h)-5) - sqrt(x-5))/h #
# \ \ = lim_(h rarr 0) (sqrt(x+h-5) - sqrt(x-5))/h *(sqrt(x+h-5) + sqrt(x-5))/(sqrt(x+h-5) + sqrt(x-5))#
# \ \ = lim_(h rarr 0) ((sqrt(x+h-5) - sqrt(x-5)) (sqrt(x+h-5) + sqrt(x-5))) / (h(sqrt(x+h-5) + sqrt(x-5)))#
# \ \ = lim_(h rarr 0) ( sqrt(x+h-5)^2 - sqrt(x-5)^2 ) / (h(sqrt(x+h-5) + sqrt(x-5)))#
# \ \ = lim_(h rarr 0) ( (x+h-5) - (x-5) ) / (h(sqrt(x+h-5) + sqrt(x-5)))#
# \ \ = lim_(h rarr 0) ( h ) / (h(sqrt(x+h-5) + sqrt(x-5)))#
# \ \ = lim_(h rarr 0) ( 1 ) / (sqrt(x+h-5) + sqrt(x-5))#
# \ \ = ( 1 ) / (sqrt(x+0-5) + sqrt(x-5))#
# \ \ = ( 1 ) / (sqrt(x-5) + sqrt(x-5))#
# \ \ = ( 1 ) / (2sqrt(x-5))#
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Answer 2

To find the derivative of the function ( \sqrt{x-5} ) using first principles, we'll use the definition of the derivative:

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

For the given function ( \sqrt{x-5} ), let's denote it as ( f(x) ).

So, ( f(x) = \sqrt{x - 5} ).

Now, we'll substitute ( f(x) ) and ( f(x + h) ) into the definition of the derivative:

[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h - 5} - \sqrt{x - 5}}{h} ]

Next, we'll rationalize the numerator by multiplying both the numerator and denominator by the conjugate of the numerator:

[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h - 5} - \sqrt{x - 5}}{h} \times \frac{\sqrt{x + h - 5} + \sqrt{x - 5}}{\sqrt{x + h - 5} + \sqrt{x - 5}} ]

Simplify the expression:

[ f'(x) = \lim_{h \to 0} \frac{(x + h - 5) - (x - 5)}{h(\sqrt{x + h - 5} + \sqrt{x - 5})} ]

[ f'(x) = \lim_{h \to 0} \frac{x + h - 5 - x + 5}{h(\sqrt{x + h - 5} + \sqrt{x - 5})} ]

[ f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h - 5} + \sqrt{x - 5})} ]

[ f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x + h - 5} + \sqrt{x - 5}} ]

Now, as ( h ) approaches 0, the expression becomes:

[ f'(x) = \frac{1}{2\sqrt{x - 5}} ]

Therefore, the derivative of ( \sqrt{x - 5} ) using first principles is ( \frac{1}{2\sqrt{x - 5}} ).

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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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