Derivative of function with radicals

[PhysChem]

I've been having trouble figuring out how to find the derivative of f(x) = x + √x

The farthest I got was:

[(x+h) + √(x+h) - (x+√x)]/h =

[h + √(x+h) - √x] / h

I got stuck here because I'm not sure how to cancel out h in numerator and denominator (if i can even do that at this stage) or multiply the whole equation by the conjugate. For some reason I'm stumped on what seems like simple algebra.

 

[HallsofIvy]

I've been having trouble figuring out how to find the derivative of f(x) = x + √x

The farthest I got was:

[(x+h) + √(x+h) - (x+√x)]/h =

[h + √(x+h) - √x] / h

I got stuck here because I'm not sure how to cancel out h in numerator and denominator (if i can even do that at this stage) or multiply the whole equation by the conjugate. For some reason I'm stumped on what seems like simple algebra.

That is, of course, 1+ \frac{\sqrt{x+h}- \sqrt{x}}{h}. The "1" is of course, the derivative of "x". All that is left is the last fraction. To do that "rationalize" the numerator: multiply both numerator and denominator by \sqrt{x+h}+ \sqrt{x} to get
\frac{\sqrt{x+h}- \sqrt{x}}{h}\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}
= \frac{(x+ h)- x}{h(\sqrt{x+h}+\sqrt{x})}= \frac{1}{\sqrt{x+h}+ \sqrt{x}}

Now, it's easy to take the limit as h goes to 0.

 

[PhysChem]

I see, thank you for the help! the "1 + (fraction)" was what I missed.

Does this mean my second step [h + √(x+h) - √x] / h was wrong?

I'm still not sure how to arrive at your first step, the "1 + (fraction)". However, everything else makes sense now!

 

[eumyang]

I see, thank you for the help! the "1 + (fraction)" was what I missed.

Does this mean my second step [h + √(x+h) - √x] / h was wrong?

I'm still not sure how to arrive at your first step, the "1 + (fraction)". However, everything else makes sense now!

No your second step isn't wrong. HallsofIvy's first step comes from your second step:
\frac{h + \sqrt{x+h}- \sqrt{x}}{h}
= \frac{h}{h} + \frac{\sqrt{x+h}- \sqrt{x}}{h}
= 1 + \frac{\sqrt{x+h}- \sqrt{x}}{h}