Using the limit definition, how do you differentiate #f(x) = x^(1/2) #?

Question

Christopher Agresta · Accepted Answer

The crucial step is #((x+h)^(1/2) - x^(1/2))/h = h/(h((x+h)^(1/2) + x^(1/2)))#

Here is the algebra of the crucial step of "rationalizing" the numerator:

#((x+h)^(1/2) - x^(1/2))/h = (((x+h)^(1/2) - x^(1/2)))/(h) (((x+h)^(1/2) + x^(1/2)))/(((x+h)^(1/2) + x^(1/2)))#

= ((x+h)-x)//(h((x+h)^(1/2) + x^(1/2)))

# = h/(h((x+h)^(1/2) + x^(1/2)))#

Using this algebra, we get:

#lim_(hrarr0) ((x+h)^(1/2) - x^(1/2))/h = lim_(hrarr0)h/(h((x+h)^(1/2) + x^(1/2)))#

# = lim_(hrarr0) 1/((x+h)^(1/2) + x^(1/2))#

# = 1/(x^(1/2)+x^(1/2)) = 1/(2x^(1/2))#

Christopher Allers · Answer

undefined To differentiate $ f(x) = x^{\frac{1}{2}} $ using the limit definition, we first write the difference quotient:

$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$

Substitute $ f(x) = x^{\frac{1}{2}} $ into the difference quotient:

$$ f'(x) = \lim_{h \to 0} \frac{(x+h)^{\frac{1}{2}} - x^{\frac{1}{2}}}{h} $$

Next, we use the binomial expansion to simplify $ (x+h)^{\frac{1}{2}} $:

$$ (x+h)^{\frac{1}{2}} = x^{\frac{1}{2}} + \frac{1}{2}h x^{-\frac{1}{2}} + O(h^2) $$

Substitute the expansion into the difference quotient:

$$ f'(x) = \lim_{h \to 0} \frac{x^{\frac{1}{2}} + \frac{1}{2}h x^{-\frac{1}{2}} + O(h^2) - x^{\frac{1}{2}}}{h} $$

Simplify:

$$ f'(x) = \lim_{h \to 0} \frac{\frac{1}{2}h x^{-\frac{1}{2}} + O(h^2)}{h} $$

$$ f'(x) = \lim_{h \to 0} \left(\frac{\frac{1}{2}h x^{-\frac{1}{2}}}{h} + \frac{O(h^2)}{h}\right) $$

$$ f'(x) = \lim_{h \to 0} \left(\frac{\frac{1}{2} x^{-\frac{1}{2}}}{1} + \frac{O(h^2)}{h}\right) $$

$$ f'(x) = \frac{1}{2} x^{-\frac{1}{2}} $$

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