How do you find f'(x) using the limit definition given #f(x)=x^(-1/2)#?

Question

Cameron Alexander · Accepted Answer

see below

By definition: #f'(x) = lim_(h to 0) (f(x+h) - f(x))/(h)# Here we have #f'(x) = lim_(h to 0) 1/h * (1/sqrt(x+h) - 1/sqrt(x))# # = lim_(h to 0) 1/h * (( sqrt(x) - sqrt(x+h) )/(sqrt(x+h) sqrt(x)))# A very common tactic when dealing with radicals is to multiply by the conjugate, and use the idea that #(a+b)(a-b) = a^2 - b^2# Here that means: # = lim_(h to 0) 1/h * (( sqrt(x) - sqrt(x+h) )/(sqrt(x+h) sqrt(x)))* (( sqrt(x) + sqrt(x+h) )/( sqrt(x) + sqrt(x+h)))# # = lim_(h to 0) 1/h * (( (x) - (x+h) )/(x sqrt(x+h) + (x+h) sqrt(x)))# doing the numerator first # = lim_(h to 0) ( -1 )/(x sqrt(x+h) + (x+h) sqrt(x))# then the denominator # = lim_(h to 0) - ( 1 )/(x sqrt(x+h) + x sqrt(x) + h sqrt(x))# # = - ( 1 )/(x sqrt(x) + x sqrt(x) + 0)# # = - ( 1 )/(2x sqrt(x) )# by some more manipulation, this then becomes the same as you would expect from a straight power rule # = - ( 1 )/(2 sqrt(x^3) )# # = - ( 1 )/(2 x^(3/2) )#

Isabella Ackerman · Answer

undefined To find $ f'(x) $ using the limit definition, where $ f(x) = x^{-1/2} $, 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 the given function $ f(x) = x^{-1/2} $ into the formula:
   $$ f'(x) = \lim_{h \to 0} \frac{(x + h)^{-1/2} - x^{-1/2}}{h} $$

3. Simplify the expression:
   $$ f'(x) = \lim_{h \to 0} \frac{(x + h)^{-1/2} - x^{-1/2}}{h} $$
   $$ = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h} $$

4. Rationalize the numerator:
   $$ f'(x) = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h\sqrt{x}\sqrt{x + h}} $$

5. Combine the terms:
   $$ f'(x) = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h\sqrt{x}\sqrt{x + h}} $$
   $$ = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h} \cdot \frac{1}{\sqrt{x}\sqrt{x + h}} $$

6. Apply the conjugate rule:
   $$ f'(x) = \lim_{h \to 0} \frac{(\sqrt{x} - \sqrt{x + h})(\sqrt{x} + \sqrt{x + h})}{h(\sqrt{x}\sqrt{x + h})(\sqrt{x} + \sqrt{x + h})} $$

7. Simplify the numerator:
   $$ f'(x) = \lim_{h \to 0} \frac{x - (x + h)}{h(\sqrt{x}\sqrt{x + h})(\sqrt{x} + \sqrt{x + h})} $$

8. Cancel out $ h $:
   $$ f'(x) = \lim_{h \to 0} \frac{-h}{h(\sqrt{x}\sqrt{x + h})(\sqrt{x} + \sqrt{x + h})} $$

9. Simplify:
   $$ f'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{x}\sqrt{x + h}(\sqrt{x} + \sqrt{x + h})} $$

10. Evaluate the limit as $ h $ approaches $ 0 $:
   $$ f'(x) = \frac{-1}{2x^{3/2}} $$

Therefore, the derivative of $ f(x) = x^{-1/2} $ is $ f'(x) = \frac{-1}{2x^{3/2}} $.