How do you use the limit definition of the derivative to find the derivative of #f(x)=sqrt(2x+7)#?
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To use the limit definition of the derivative to find the derivative of ( f(x) = \sqrt{2x + 7} ), follow these steps:

Start with the limit definition of the derivative: [ f'(x) = \lim_{{h \to 0}} \frac{f(x + h)  f(x)}{h} ]

Substitute the given function ( f(x) = \sqrt{2x + 7} ) into the formula: [ f'(x) = \lim_{{h \to 0}} \frac{\sqrt{2(x + h) + 7}  \sqrt{2x + 7}}{h} ]

Rationalize the numerator by multiplying the numerator and denominator by the conjugate of the numerator: [ f'(x) = \lim_{{h \to 0}} \frac{\sqrt{2(x + h) + 7}  \sqrt{2x + 7}}{h} \cdot \frac{\sqrt{2(x + h) + 7} + \sqrt{2x + 7}}{\sqrt{2(x + h) + 7} + \sqrt{2x + 7}} ]

Simplify the numerator and denominator: [ f'(x) = \lim_{{h \to 0}} \frac{(2(x + h) + 7)  (2x + 7)}{h(\sqrt{2(x + h) + 7} + \sqrt{2x + 7})} ]

Expand and simplify the numerator: [ f'(x) = \lim_{{h \to 0}} \frac{2x + 2h + 7  2x  7}{h(\sqrt{2(x + h) + 7} + \sqrt{2x + 7})} ] [ f'(x) = \lim_{{h \to 0}} \frac{2h}{h(\sqrt{2(x + h) + 7} + \sqrt{2x + 7})} ]

Cancel out ( h ) in the numerator and denominator: [ f'(x) = \lim_{{h \to 0}} \frac{2}{\sqrt{2(x + h) + 7} + \sqrt{2x + 7}} ]

Evaluate the limit as ( h ) approaches 0: [ f'(x) = \frac{2}{\sqrt{2x + 7} + \sqrt{2x + 7}} ] [ f'(x) = \frac{2}{2\sqrt{2x + 7}} ] [ f'(x) = \frac{1}{\sqrt{2x + 7}} ]
So, the derivative of ( f(x) = \sqrt{2x + 7} ) is ( f'(x) = \frac{1}{\sqrt{2x + 7}} ).
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When evaluating a onesided 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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