How do you use the definition of a derivative to find the derivative of #f(x) = x + sqrtx#?
The definition of a derivative is
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To find the derivative of ( f(x) = x + \sqrt{x} ), we can use the definition of a derivative, which states that the derivative of a function ( f(x) ) at a point ( x ) is equal to the limit as ( h ) approaches 0 of ( \frac{f(x + h) - f(x)}{h} ).
So, let's apply this definition to ( f(x) = x + \sqrt{x} ):
[ \begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \ &= \lim_{h \to 0} \frac{(x + h) + \sqrt{x + h} - (x + \sqrt{x})}{h} \ &= \lim_{h \to 0} \frac{x + h - x}{h} + \frac{\sqrt{x + h} - \sqrt{x}}{h} \ &= \lim_{h \to 0} \frac{h}{h} + \frac{\sqrt{x + h} - \sqrt{x}}{h} \ &= \lim_{h \to 0} 1 + \frac{\sqrt{x + h} - \sqrt{x}}{h} \end{align*} ]
To proceed, we'll rationalize the numerator by multiplying and dividing by the conjugate of the expression:
[ \begin{align*} f'(x) &= \lim_{h \to 0} \left(1 + \frac{\sqrt{x + h} - \sqrt{x}}{h}\right) \ &= \lim_{h \to 0} \left(1 + \frac{(\sqrt{x + h} - \sqrt{x})(\sqrt{x + h} + \sqrt{x})}{h(\sqrt{x + h} + \sqrt{x})}\right) \ &= \lim_{h \to 0} \left(1 + \frac{x + h - x}{h(\sqrt{x + h} + \sqrt{x})}\right) \ &= \lim_{h \to 0} \left(1 + \frac{h}{h(\sqrt{x + h} + \sqrt{x})}\right) \ &= \lim_{h \to 0} \left(1 + \frac{1}{\sqrt{x + h} + \sqrt{x}}\right) \ &= 1 + \frac{1}{2\sqrt{x}} \ &= \frac{1}{2\sqrt{x}} + 1 \end{align*} ]
So, ( f'(x) = \frac{1}{2\sqrt{x}} + 1 ).
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To use the definition of a derivative to find the derivative of ( f(x) = x + \sqrt{x} ), we first recall the definition of a derivative:
[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]
Now, let's apply this definition to ( f(x) = x + \sqrt{x} ):
[ f'(x) = \lim_{h \to 0} \frac{(x + h) + \sqrt{x + h} - (x + \sqrt{x})}{h} ]
[ f'(x) = \lim_{h \to 0} \frac{x + h + \sqrt{x + h} - x - \sqrt{x}}{h} ]
[ f'(x) = \lim_{h \to 0} \frac{h + \sqrt{x + h} - \sqrt{x}}{h} ]
To simplify this expression, we'll rationalize the numerator by multiplying by the conjugate:
[ f'(x) = \lim_{h \to 0} \frac{(h + \sqrt{x + h} - \sqrt{x})(\sqrt{x + h} + \sqrt{x})}{h(\sqrt{x + h} + \sqrt{x})} ]
[ f'(x) = \lim_{h \to 0} \frac{h(\sqrt{x + h} + \sqrt{x}) + (x + h - x)}{h(\sqrt{x + h} + \sqrt{x})} ]
[ f'(x) = \lim_{h \to 0} \frac{h\sqrt{x + h} + h\sqrt{x} + x + h - x}{h(\sqrt{x + h} + \sqrt{x})} ]
[ f'(x) = \lim_{h \to 0} \frac{h\sqrt{x + h} + h\sqrt{x} + h}{h(\sqrt{x + h} + \sqrt{x})} ]
[ f'(x) = \lim_{h \to 0} \frac{h(\sqrt{x + h} + \sqrt{x} + 1)}{h(\sqrt{x + h} + \sqrt{x})} ]
Now, we can cancel out the ( h ) terms:
[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h} + \sqrt{x} + 1}{\sqrt{x + h} + \sqrt{x}} ]
[ f'(x) = \frac{\sqrt{x} + \sqrt{x} + 1}{\sqrt{x} + \sqrt{x}} ]
[ f'(x) = \frac{2\sqrt{x} + 1}{2\sqrt{x}} ]
[ f'(x) = \frac{1}{2\sqrt{x}} + \frac{1}{2} ]
So, the derivative of ( f(x) = x + \sqrt{x} ) is ( f'(x) = \frac{1}{2\sqrt{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.
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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