Troubleshooting Differentiation Using the Long Formula

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Homework Statement

I have to differentiate using the long formula [f(x+h)-f(x)] / (h).


Homework Equations



f(x) = (1)/(2+x), x=2

The Attempt at a Solution



First I wrote [ (1/2+x+h) - (1/2+x) ] / (h). Then I created a common denominator in the numerator. I then made the bottom denominator a fraction.

Code:
(2+x)-(2+x+h)     (1)
___________    x  __
(2+x+h)(2+x)      (h)

I subtracted common variables from the numerator and got.

Code:
   -(h)                (1)
________             x ___ 
[(2+x+h)(2+x)]         (h)

Then I got:

       -(1)
_______________
(2+x+h)(2+x)

But as I expanded the numerator, I found that I couldn't get rid of the last h. What is the problem?

Thank you.
 
Last edited:
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i think you dropped a minus sign

also remember h is tending to zero... so multiple out, then take the limit & you can cancel terms tending to zero
 
note you can write in tex (click on below to see the code)
\frac{-1}{(2+x)(2+x+h)} = \frac{-1}{(2+x)^2+h(2+x))}
 
Thanks for the quick reply.

That would mean the solution to this particular problem is - \frac{1}{16}?
 
Yes.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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