How do you find f'(x) using the definition of a derivative for #f(x)=sqrt(2x-1)#?
See the explanation.
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To find ( f'(x) ) using the definition of a derivative for ( f(x) = \sqrt{2x - 1} ), we use the definition of the derivative:
[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]
Substitute the function ( f(x) = \sqrt{2x - 1} ) into the definition:
[ f'(x) = \lim_{h \to 0} \frac{\sqrt{2(x + h) - 1} - \sqrt{2x - 1}}{h} ]
To simplify this expression, we use the conjugate method to rationalize the numerator:
[ f'(x) = \lim_{h \to 0} \frac{\sqrt{2(x + h) - 1} - \sqrt{2x - 1}}{h} \times \frac{\sqrt{2(x + h) - 1} + \sqrt{2x - 1}}{\sqrt{2(x + h) - 1} + \sqrt{2x - 1}} ]
[ f'(x) = \lim_{h \to 0} \frac{(2(x + h) - 1) - (2x - 1)}{h(\sqrt{2(x + h) - 1} + \sqrt{2x - 1})} ]
[ f'(x) = \lim_{h \to 0} \frac{2x + 2h - 1 - 2x + 1}{h(\sqrt{2(x + h) - 1} + \sqrt{2x - 1})} ]
[ f'(x) = \lim_{h \to 0} \frac{2h}{h(\sqrt{2(x + h) - 1} + \sqrt{2x - 1})} ]
[ f'(x) = \lim_{h \to 0} \frac{2}{\sqrt{2(x + h) - 1} + \sqrt{2x - 1}} ]
[ f'(x) = \frac{2}{\sqrt{2x - 1} + \sqrt{2x - 1}} ]
[ f'(x) = \frac{2}{2\sqrt{2x - 1}} ]
[ f'(x) = \frac{1}{\sqrt{2x - 1}} ]
So, ( f'(x) = \frac{1}{\sqrt{2x - 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.
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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