Derivative of f(x) = x + sqrt(x)

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January 31st 2011, 08:35 PM
joatmon

Derivative of f(x) = x + sqrt(x)

Find f'(x) for the function $f(x) = x + \sqrt{x}$

This is Calc 1, so we have to do this the long way for a little while longer, so I have to use the formula

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

I can't seem to get this one going. I can easily simplify it down to:

$\displaystyle \lim_{h\to 0 } \frac{h + \sqrt{x+h} - \sqrt{x}}{h}$

But then I don't know where to take it. There is no easy conjugate to multiply by, and I don't see any way to factor the numerator. Can anybody give me a hint about what operation to try on this thing to move it along?

Thanks.
January 31st 2011, 08:52 PM
CaptainBlack

Quote:

Originally Posted by joatmon

Find f'(x) for the function $f(x) = x + \sqrt{x}$

This is Calc 1, so we have to do this the long way for a little while longer, so I have to use the formula

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

I can't seem to get this one going. I can easily simplify it down to:

$\displaystyle \lim_{h\to 0 } \frac{h + \sqrt{x+h} - \sqrt{x}}{h}$

But then I don't know where to take it. There is no easy conjugate to multiply by, and I don't see any way to factor the numerator. Can anybody give me a hint about what operation to try on this thing to move it along?

Thanks.

$\displaystyle \lim_{h\to 0 } \frac{h + \sqrt{x+h} - \sqrt{x}}{h}=1+ \lim_{h\to 0 } \frac{ \sqrt{x+h} - \sqrt{x}}{h}$

Now:

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

....

CB
January 31st 2011, 09:50 PM
joatmon

Sure. That will work. Thanks a lot. I appreciate it.