Find dy/dx of radical expression

[QuarkCharmer]

Homework Statement

$$\frac{d}{dx} x+\sqrt{x}$$
Using the definition of a derivative, no rules allowed.

Homework Equations

The Attempt at a Solution

$$\frac{d}{dx} x+\sqrt{x}$$
$$\lim_{h\to0} \frac{(x+h)+\sqrt{x+h}-x-\sqrt{x}}{h}$$
$$\lim_{h\to0} \frac{(x+h)+\sqrt{x+h}-x-\sqrt{x}}{h}$$
$$\lim_{h\to0} \frac{h+\sqrt{x+h}-\sqrt{x}}{h}$$

Not sure how to go about canceling out that h in the denominator before taking the limit. If there were two expressions on the top, then multiplying in a conjugate usually works, but I'm not sure what to do next?

My guess is to arrange it so that it's 1+(root(x+h)-root(x))/h and then put the conjugate into the right side. Trying that now

 

[micromass]

Multiply the numerator and denominator with

$$h+\sqrt{x+h}+\sqrt{x}$$

 

[QuarkCharmer]

Ah, I was thinking that would only fork for a binomial. I'll give that a shot if this doesn't work out.

 

[micromass]

Well, this is a binomial: [itex](h+\sqrt{x+h})-\sqrt{x}[/tex]... <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f600.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":biggrin:" title="Big Grin :biggrin:" data-smilie="8"data-shortname=":biggrin:" />[/itex]

 

[QuarkCharmer]

Yeah, I see that now. Lesson learned.

 

[lanedance]

I would consider the radical part seprately
$$\lim_{h\to0} \frac{h+\sqrt{x+h}-\sqrt{x}}{h}$$

$$\lim_{h\to0} \frac{h}{h} + \frac{\sqrt{x+h}-\sqrt{x}}{h}$$

then multiply the radical part by
$$1= \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$$

this is the same as calculating the derivative fo x & sqrt(x) separately

 

[QuarkCharmer]

Okay, I am trying to handle it the 1+blah/blah method.

$$\lim_{h\to0} \frac{h+\sqrt{x+h}-\sqrt{x}}{h}$$

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

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

$$\lim_{h\to0} 1+ \frac{h}{h(\sqrt{x+h}+\sqrt{x})}$$

$$\lim_{h\to0} 1+ \frac{1}{(\sqrt{x+h}+\sqrt{x})}$$

So now I am guessing it's about time to take the limit,
Making the derivative something like:
$$f'(x) = 1+ \frac{1}{2\sqrt{x}}$$

Look good?

 

[Char. Limit]

Looks good to me.

 

[micromass]

This is perfect, quark!

Just, watch out with your notation:

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

 

[QuarkCharmer]

Ah, forgot a parenthesis there. On my paper it's correct! I am horrible with latex once the lines get longer than 20 characters.

Thanks for the help yet again!