How do you find the derivative of # y=sqrt(x−3)# using the limit definition?

Question

Isabella Ackerman · Accepted Answer

Given: #f(x) = y = sqrt(x−3)#

Then:

#f(x+h) = sqrt(x+h−3)#

Using the limit definition:

#f'(x) = lim_(h to 0) (f(x+h)-f(x))/h#

Substitute in the functions:

#f'(x) = lim_(h to 0) (sqrt(x+h−3)-sqrt(x−3))/h#

We know that, if we multiply the numerator by #sqrt(x+h−3)+sqrt(x−3)#, we will eliminate the radicals but we must, also, multiply the denominator by the same thing:

#f'(x) = lim_(h to 0) (sqrt(x+h−3)+sqrt(x−3))/(sqrt(x+h−3)+sqrt(x−3))(sqrt(x+h−3)-sqrt(x−3))/h#

Please observe that we have not changed the value of anything, because we have multiplied by 1 in a very special form.

Now, the radicals in the numerator disappear, because of the property #(a+b)(a-b) = a^2-b^2#, and the denominators just multiply:

#f'(x) = lim_(h to 0) ((x+h−3)-(x−3))/(h(sqrt(x+h−3)+sqrt(x−3)))#

Combine like terms in the numerator:

#f'(x) = lim_(h to 0) h/(h(sqrt(x+h−3)+sqrt(x−3)))#

The h in the numerator cancels with the h in the denominator:

#f'(x) = lim_(h to 0) 1/((sqrt(x+h−3)+sqrt(x−3)))#

Now we can allow h to become 0:

#f'(x) = 1/((sqrt(x−3)+sqrt(x−3)))#

Combine like terms:

#f'(x) = 1/(2sqrt(x−3))#

Christopher Agresta · Answer

undefined To find the derivative of $ y = \sqrt{x - 3} $ using the limit definition, we apply the definition of the derivative:

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

Substitute $ f(x) = \sqrt{x - 3} $ into the definition and simplify the expression:

$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h - 3} - \sqrt{x - 3}}{h} $$

To eliminate the square roots in the numerator, we rationalize the expression by multiplying the numerator and denominator by the conjugate of the numerator:

$$ f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h - 3} - \sqrt{x - 3}}{h} \cdot \frac{\sqrt{x + h - 3} + \sqrt{x - 3}}{\sqrt{x + h - 3} + \sqrt{x - 3}} $$

$$ f'(x) = \lim_{h \to 0} \frac{(x + h - 3) - (x - 3)}{h(\sqrt{x + h - 3} + \sqrt{x - 3})} $$

$$ f'(x) = \lim_{h \to 0} \frac{x + h - 3 - x + 3}{h(\sqrt{x + h - 3} + \sqrt{x - 3})} $$

$$ f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x + h - 3} + \sqrt{x - 3})} $$

$$ f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x + h - 3} + \sqrt{x - 3}} $$

Now, as $ h $ approaches 0, we evaluate the limit:

$$ f'(x) = \frac{1}{2\sqrt{x - 3}} $$

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