Using the limit definition, how do you find the derivative of #f (x) = sqrt (x−3)#?

Question

Paisley Johnson · Accepted Answer

I found: #f'(x)=1/(2sqrt(x-3))#

Have a look:

Adam Adkins · Answer

undefined To find the derivative of the function $ f(x) = \sqrt{x - 3} $ using the limit definition of the derivative, follow these steps:

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

2. Substitute the given function into the definition:
   $$ f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h - 3} - \sqrt{x - 3}}{h} $$

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

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

5. Simplify further:
   $$ f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h - 3} + \sqrt{x - 3})} $$

6. Cancel out $ h $ in the numerator and denominator:
   $$ f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x + h - 3} + \sqrt{x - 3}} $$

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

So, the derivative of $ f(x) = \sqrt{x - 3} $ using the limit definition is $ f'(x) = \frac{1}{2\sqrt{x - 3}} $.