Calculating derivative using regular difference quotient

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Vertciel

Junior Member
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May 13, 2007
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78
Hello there,

I am self-studying calculus using The Complete Idiot's Guide to Calculus, and there is a problem asking me to calculate the derivative of a function with the alternate difference quotient.

However, for extra practise, I am trying to find the derivative of a function with the regular difference quotient. I am having trouble on one step so I would appreciate any help.

Thank you.

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1. Calculate the derivative f(x)=x+1\displaystyle f(x) = \sqrt{x + 1}.

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lim⁡Δx→0f(x+Δx)−f(x)Δx\displaystyle \lim_{\Delta x \to 0}\frac {f(x + \Delta x) - f(x)}{\Delta x}

Determining f(x+Δx)\displaystyle f(x + \Delta x):

f(x+Δx)=(x+Δx)+1\displaystyle f(x + \Delta x) = \sqrt{(x + \Delta x) + 1}

Therefore:

lim⁡Δx→0f(x+Δx)−f(x)Δx\displaystyle \lim_{\Delta x \to 0}\frac {f(x + \Delta x) - f(x)}{\Delta x}

= lim⁡Δx→0x+Δx+1−x+1Δx\displaystyle \lim_{\Delta x \to 0}\frac {\sqrt{x + \Delta x + 1} - \sqrt{x + 1}}{\Delta x}

How would I eliminate the square roots to proceed with the algebra?
 
Re: Calculating derivative

rationalize the numerator.

multiply the last expression by

x+Δx+1+x+1x+Δx+1+x+1\displaystyle \frac{\sqrt{x + \Delta x + 1} + \sqrt{x + 1}}{\sqrt{x + \Delta x + 1} + \sqrt{x + 1}},

simplify, then take the limit.
 
Re: Calculating derivative

Thanks for the help! I forgot about multiplying the fraction by its conjugate.
 
Is there?. I was just kidding. That IS amazing. I would like to see that.
 
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