Help with derivative

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June 28th 2012, 04:56 PM
rortiz81

Help with derivative

I need to find the derivative of $f(x)=x+\sqrt x+3$

I have gotten this far: $f'(x)=lim_{h\to0}\frac{f(x+h)-f(x)}{h}=lim_{h\to0}\frac{(x+h)+\sqrt{x+h} +3-\left(x+\sqrt x +3\right)}{h}=lim_{h\to0}\frac{h+\sqrt{x+h}-\sqrt x}{h}$

At this point I have tried some crazy stuff to eliminate the h on the bottom. The closest I have come is this: $lim_{h\to0}\frac{h}{h}+\frac{\sqrt{x+h}}{h}-\frac{\sqrt x}{h}$ which can be broken down into, $lim_{h\to0}\frac{h}{h}+\frac{\sqrt x}{h}+\frac{\sqrt h}{h}-\frac{\sqrt x}{h}$ which leaves this: $lim_{h\to0}\frac{h}{h}+\frac{\sqrt h}{h}$ , right?

If I haven't confused myself along the way, then I should be on track, and the answer seems tantilizingly close, but I get $lim_{h\to0}1+\frac{\sqrt h}{h}$ = $lim_{h\to0}1+\frac{1}{\sqrt h}$

What am I doing wrong?
June 28th 2012, 05:03 PM
Cbarker1

Re: Help with derivative

The step that you did incorrectly was the \sqrt((x+h))/h not (\sqrt(x)/h-(\sqrt(h)/h))
June 28th 2012, 05:42 PM
Reckoner

Re: Help with derivative

Quote:

Originally Posted by rortiz81

$\cdots+\frac{\sqrt{x+h}}{h}+\cdots$ which can be broken down into, $\cdots+\frac{\sqrt x}{h}+\frac{\sqrt h}{h}+\cdots$

Absolutely not. Do not do this.

$\sqrt{a + b} \neq \sqrt a + \sqrt b$

Quote:

Originally Posted by rortiz81

I need to find the derivative of $f(x)=x+\sqrt x+3$

Rationalizing the numerator can be a good strategy. Continuing your work from before you tried to split everything up:

$f'(x) = \lim_{h\to0}\frac{h + \sqrt{x + h} - \sqrt x}h$

$=\lim_{h\to0}\left(1 + \frac{\sqrt{x+h}-\sqrt x}h\right)$

$=\lim_{h\to0}\left(1 + \frac{\sqrt{x+h}-\sqrt x}h\cdot\frac{\sqrt{x+h}+\sqrt x}{\sqrt{x+h}+\sqrt x}\right)$

$=\lim_{h\to0}\left[1 + \frac{(x + h) - x}{h\left(\sqrt{x+h} + \sqrt x\right)}\right]$

$=\lim_{h\to0}\left[1 + \frac h{h\left(\sqrt{x+h} + \sqrt x\right)}\right]$

$=\lim_{h\to0}\left(1 + \frac1{\sqrt{x+h} + \sqrt x}\right)$

$=1+\frac1{2\sqrt{x}}$
June 28th 2012, 05:52 PM
richard1234

Re: Help with derivative

If you can prove/use the fact that $\frac{d}{dx} (f(x) + g(x)) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)$ for differentiable functions f and g, then the derivative is equal to

$\frac{d}{dx} x + \frac{d}{dx} \sqrt{x} - \frac{d}{dx} 3$ , which is equal to

$1 + \frac{1}{2 \sqrt{x}}$
June 28th 2012, 06:16 PM
rortiz81

Re: Help with derivative

Thank you all. Ugh. I really should have better algebra skills than this. BTW, this forum rocks.