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October 3rd, 2017, 10:51 AM   #1
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Derivative causing div-by-zero problem

I can figure out the 1st derivative of this by using the quotient rule, but if I try to use the definition of the derivative to figure it out, I can't escape running into a division-by-zero. Can anyone show me what algebra trick(s) I'm missing?
The function is f(x) =$\displaystyle \frac{2}{\sqrt{3-x}}$, so...

f'(x) = $\displaystyle \lim_{h\rightarrow 0}1/h \left [ \frac{2}{\sqrt{3-(x+h)}} - \frac{2}{\sqrt{3-x}}\right ]$

No matter how I manipulate this expression (many sheets of paper), I always wind up with 'h' as a factor in some denominator.
Can someone show me a way to solve this?
Thanks.
Uncle Ed
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October 3rd, 2017, 06:20 PM   #2
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After combining the fractions, multiplying the numerator and denominator by $\sqrt{3-x} + \sqrt{3-(x+h)}$ causes the numerator to become $2h$, allowing $h$ to be cancelled.
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October 4th, 2017, 05:55 AM   #3
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Thank you!

I thought I had tried using the conjugate but must have made an algebra mistake. This time it worked!
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October 4th, 2017, 03:53 PM   #4
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Amazing, isn't it?

I once made the mistake of agreeing to teach a course in "Calculus for Business Administration". The text was assigned by the Department of Business Administration.

On one page the "laws of limits", "the limit of a sum is the sum of the limits", "the limit of a product is the product of the limits", and "the limit of a quotient is the quotient of the limits", with out proof, on one day. Given the course, I didn't mind that no proofs were given but on the very next page, it defined the derivative as the limit of the difference quotient, completely ignoring the fact that, since the difference quotient necessarily has denominator going to 0, none of those limit laws apply!
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