How do you find f'(x) using the definition of a derivative #f(x) = sqrtx + 2#?

Question

Evelyn Agard · Accepted Answer

I found: #1/(2sqrt(x))#

Have a look:

Cameron Alexander · Answer

undefined To find $ f'(x) $ using the definition of a derivative for the function $ f(x) = \sqrt{x} + 2 $, we first recall the definition:

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

Now, we substitute $ f(x) = \sqrt{x} + 2 $ into the definition:

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

Next, we simplify the expression:

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

To eliminate the square roots from the numerator, we multiply the expression by the conjugate of the numerator:

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

This simplifies to:

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

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

Now, we can evaluate the limit:

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

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

So, the derivative of $ f(x) = \sqrt{x} + 2 $ is $ f'(x) = \frac{1}{{2\sqrt{x}}} $.