# Help with derivative

• Jun 28th 2012, 04:56 PM
rortiz81
Help with derivative
I need to find the derivative of $\displaystyle f(x)=x+\sqrt x+3$

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

At this point I have tried some crazy stuff to eliminate the h on the bottom. The closest I have come is this: $\displaystyle lim_{h\to0}\frac{h}{h}+\frac{\sqrt{x+h}}{h}-\frac{\sqrt x}{h}$ which can be broken down into, $\displaystyle lim_{h\to0}\frac{h}{h}+\frac{\sqrt x}{h}+\frac{\sqrt h}{h}-\frac{\sqrt x}{h}$ which leaves this:$\displaystyle lim_{h\to0}\frac{h}{h}+\frac{\sqrt h}{h}$, 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 $\displaystyle lim_{h\to0}1+\frac{\sqrt h}{h}$=$\displaystyle lim_{h\to0}1+\frac{1}{\sqrt h}$

What am I doing wrong?
• Jun 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))
• Jun 28th 2012, 05:42 PM
Reckoner
Re: Help with derivative
Quote:

Originally Posted by rortiz81
$\displaystyle \cdots+\frac{\sqrt{x+h}}{h}+\cdots$ which can be broken down into, $\displaystyle \cdots+\frac{\sqrt x}{h}+\frac{\sqrt h}{h}+\cdots$

Absolutely not. Do not do this.

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

Quote:

Originally Posted by rortiz81
I need to find the derivative of $\displaystyle f(x)=x+\sqrt x+3$

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

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

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

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

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

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

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

$\displaystyle =1+\frac1{2\sqrt{x}}$
• Jun 28th 2012, 05:52 PM
richard1234
Re: Help with derivative
If you can prove/use the fact that $\displaystyle \frac{d}{dx} (f(x) + g(x)) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)$ for differentiable functions f and g, then the derivative is equal to

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

$\displaystyle 1 + \frac{1}{2 \sqrt{x}}$
• Jun 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.