How do you find the derivative of # f(x) = sqrtx# using the formal definition?

Question

Charles Arocha · Accepted Answer

I would use a "strange" rationalization of the numerator!

From the definition of derivative: #lim_(h->0)((sqrt(x+h)-sqrt(x))/h)=# I would try to rationalize the numerator (although it seems strange...). #=lim_(h->0)((sqrt(x+h)-sqrt(x))/h)*color(red)(((sqrt(x+h)+sqrt(x)))/((sqrt(x+h)+sqrt(x))))=# #=lim_(h->0)(x+h-x)/(h*(sqrt(x+h)+sqrt(x)))=# #=lim_(h->0)(h)/(h*(sqrt(x+h)+sqrt(x)))=# #=lim_(h->0)(1)/((sqrt(x+h)+sqrt(x)))=# as #h->0# #=1/(2sqrt(x))#

Evelyn Agard · Answer

Use: #f'(a) = lim_(h->0) ((f(a+h) - f(a))/h)#

to find #f'(x) = 1/(2sqrt(x))# #f'(a) = lim_(h->0) ((f(a+h) - f(a))/h)# #=lim_(h->0) ((sqrt(a+h) - sqrt(a))/h)# #=lim_(h->0) (((sqrt(a+h) - sqrt(a))(sqrt(a+h) + sqrt(a)))/(h(sqrt(a+h)+sqrt(a))))# #=lim_(h->0) (((color(red)(cancel(color(black)(a)))+h)-color(red)(cancel(color(black)(a))))/(h(sqrt(a+h)+sqrt(a))))# #=lim_(h->0) (1/(sqrt(a+h)+sqrt(a)))# #=1/(2sqrt(a))# That is: #f'(x) = 1/(2sqrt(x))#

Charles Arocha · Answer

undefined To find the derivative of $ f(x) = \sqrt{x} $ using the formal definition of a derivative, we use the definition of the derivative:

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

Substituting $ f(x) = \sqrt{x} $ into the definition:

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

To simplify this expression, we multiply the numerator and denominator by the conjugate of the numerator:

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

This simplifies to:

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

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

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

Now, substitute $ h = 0 $ into the expression:

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