I don't understand when
I apply the difference quotient to: $f(x) = \sqrt{x} $ , to get:
$$\frac{\sqrt{x+h} - \sqrt{x}}{h}$$
To simplify it.. How does it end up like this?:
$$\frac{x + h - x}{h \sqrt{x+h} + \sqrt{x}}= \frac{1}{\ 2\sqrt{x}}$$
How do the sqrt's work when moving them from the numerator to the denominator?
Thanks.
$\endgroup$ 93 Answers
$\begingroup$$\frac{\sqrt{x+h}-\sqrt{x}}{h}=\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{\sqrt{x+h}+\sqrt{x}}$
So by letting h go to 0 we get
$\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\frac{1}{2\sqrt{x}}$
$\endgroup$ 3 $\begingroup$You don't necessarily want the square root in the denominator, you want h out of it because it'll be set to 0 to find the limit. Since you can't divide by 0 you must factor h out of the denominator.
$\endgroup$ $\begingroup$actually when you want to get $ \frac{\sqrt {x+h}-\sqrt x)} h$ you don't use the limit it goes like this:
- multiply the nominator and denominator by $\sqrt{x+h}+\sqrt x$
which at the end will result in $\frac 1 {\sqrt {x+h} + \sqrt x}$
- $h$ shouldn't be treated as approaching zero unless you were told it is OR you're finding the derivative by definition.
Let me know if it helped ya :)
$\endgroup$ 1
