Deriving the Difference Quotient for a Square Root Function

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In summary, the conversation is about showing that P(x+h)-P(x)=h/[(x+h)^(1/2)+ x^(1/2)] and how to approach solving it. The suggested method is to first form P(x+h) by replacing x with x+h, then subtract P(x) from it. It is also suggested to multiply by \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} to get the desired form.
  • #1
banana_banana
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Homework Statement



If P(x)=x^(1/2)
show that P(x+h)-P(x)=h/[(x+h)^(1/2)+ x^(1/2)]

Homework Equations





The Attempt at a Solution




pls help me. I don't have any idea of this...
 
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  • #2
Try multiplying by:

[tex]\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}[/tex]
 
  • #3
Cyosis said:
Try multiplying by:

[tex]\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}[/tex]

Will i going to substitute it on the x variable? I don't know where it needs to be multiplied.
 
  • #4
"Multiply" doesn't mean substitute!

First form P(x+h) by replacing x with x+ h. Then subtract P(x) from that. That's what "P(x+h)- P(x)" means! Cyanosis is suggesting that you can get the final form you want by multiplying by
[tex]\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}[/tex]
 
  • #5
tnx for the explanation. :)
 

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