How do you use the definition of a derivative to find the derivative of #f(x) = x + sqrtx#?

Answer 1

#f'(x) = 1 + 1/(2sqrt(x))#

The definition of a derivative is

#f'(x) = lim_(h->0) (f(x+h) - f(x))/h #
# = lim_(h->0) ((x+h + sqrt(x+h)) - (x + sqrt(x)))/h#
# = lim_(h->0) (cancel(x) + h + sqrt(x + h) - cancel(x) - sqrt(x))/h#
# = lim_(h->0) (h + sqrt(x + h) - sqrt(x))/h#
# = lim_(h->0) (h/h + (sqrt(x + h) - sqrt(x))/h)#
# = lim_(h->0) 1 + lim_(h->0) (sqrt(x + h) - sqrt(x))/h#
... expand the fraction so that you can use the formula #(a+b)(a-b) = a^2 - b^2# afterwards...
# = 1 + lim_(h->0) ((sqrt(x + h) - sqrt(x))* color(blue)((sqrt(x + h) + sqrt(x))))/(h * color(blue)((sqrt(x + h) + sqrt(x))))#
# = 1 + lim_(h->0) " "((sqrt(x+h))^2 - (sqrt(x))^2)/(h * (sqrt(x + h) + sqrt(x)))#
# = 1 + lim_(h->0) " "(x + h - x)/(h * (sqrt(x + h) + sqrt(x)))#
# = 1 + lim_(h->0) " "cancel(h)/(cancel(h) * (sqrt(x + h) + sqrt(x)))#
# = 1 + lim_(h->0) " "1/(sqrt(x + h) + sqrt(x))#
At this point, you can safely plug #h = 0# to compute the limit:
# = 1 + 1/(sqrt(x) + sqrt(x))#
# = 1 + 1/(2sqrt(x))#

Hope that this helped!

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Answer 2

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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Answer 3

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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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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