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

Question

Elijah Abram · Accepted Answer

#f'(x) = 1/(2sqrt(x))# #f'(x) = lim_(Deltaxrarr0) (f(x+Deltax) - f(x))/(Deltax)# #f'(x) = lim_(Deltaxrarr0) (sqrt(x+Deltax) - sqrt(x))/(Deltax)# #f'(x) = lim_(Deltaxrarr0) (sqrt(x+Deltax) - sqrt(x))/(Deltax) * (sqrt(x+Deltax) + sqrt(x))/(sqrt(x+Deltax) + sqrt(x))# #f'(x) = lim_(Deltaxrarr0) (cancel(Deltax))/(cancel(Deltax)(sqrt(x+Deltax) + sqrt(x))# #f'(x) = lim_(Deltaxrarr0) 1/(sqrt(x+Deltax) + sqrt(x)) = 1/(2sqrt(x))#

Ezekiel Alexander · Answer

undefined To use the limit definition of the derivative to find the derivative of $f(x) = \sqrt{x}$, follow these steps:

1. Start with the definition of the derivative:
   $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

2. Substitute $f(x) = \sqrt{x}$ into the formula:
   $$f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$$

3. Rationalize the numerator by multiplying both the numerator and denominator by the conjugate of the numerator:
   $$\begin{split}& f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} \
   & = \lim_{h \to 0} \frac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x})} \
   & = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})} \
   & = \lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}\end{split}$$

4. As $h$ approaches 0, the expression becomes:
   $$f'(x) = \frac{1}{2\sqrt{x}}$$

Therefore, the derivative of $f(x) = \sqrt{x}$ is $f'(x) = \frac{1}{2\sqrt{x}}$.