# exam problems (calculus 1)

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• December 14th 2010, 08:52 AM
eq123
exam problems (calculus 1)
hey everyone, i would appreciate any help with these problems ..
• December 14th 2010, 09:09 AM
chisigma
18. Is...

$\displaystyle \frac{1}{4} \frac{x-2}{x+2} = \frac{1}{4} - \frac{1}{x+2}$ (1)

... so that...

$\displaystyle f^{(n)} (x) = (-1)^{n-1} \frac{n!}{(x+2)^{n}}$ (2)

http://digilander.libero.it/luposaba...ato&#91;1].jpg

Merry Christmas from Italy

$\chi$ $\sigma$
• December 14th 2010, 09:16 AM
Mondreus
12) Rewrite it as $\ln\left[F(x)\right] = \ln\frac{\left[f(x) \right]^\pi}{\left[3+f(x)\right]^e}=\pi \ln\left[f(x) \right]-e \ln\left[3+f(x)\right]$
• December 14th 2010, 09:20 AM
chisigma
20. Is [l'Hopital's rule...] ...

$\displaystyle \lim_{x \rightarrow 0} \frac{1-\cos 4 x}{x^{2}} = \lim_{x \rightarrow 0} \frac{4\ \sin 4x}{2 x} = \lim_{x \rightarrow 0} \frac{16\ \cos 4 x}{2} = 8$ (1)

... so that...

$\displaystyle \frac{8}{3}\ \alpha = \frac{4}{\beta} \implies \alpha\ \beta= \frac{3}{2}$ (2)

http://digilander.libero.it/luposaba...ato&#91;1].jpg

Merry Christmas from Italy

$\chi$ $\sigma$
• December 15th 2010, 01:17 AM
chisigma
12. Is...

$\displaystyle \frac{d}{dx} \ln F(x)= \frac{F^{'}(x)}{F(x)}$ (1)

... so that, writing $\displaystyle F(x)= \frac{f(x)}{g(x)}$, is also...

$\displaystyle F^{'} (x) = F(x)\ \frac{d}{dx} \ln F(x) = F(x)\ \frac{d}{dx} \{\ln f(x) - \ln g(x) \}=$

$\displaystyle = F(x)\ \frac{d}{dx} \{\frac{f^{'}(x)}{f(x)} - \frac{g^{'}(x)}{g(x)} \}$ (2)

... and now You have all the elements to compute $F^{'} (0)$ ...

http://digilander.libero.it/luposaba...ato&#91;1].jpg

Merry Christmas from Italy

$\chi$ $\sigma$
• December 15th 2010, 03:23 AM
CaptainBlack
Don't just post a list of questions like this.

1. There are too many questions for one thread, the limit is 2 or 3. I have closed this thread, repost the remaining questions in new threads with one or two questions per thread.

2. Please show some working or explain where you are having difficulties.

We are not here to do your work (I will assme this is not a take home exam) for you but to help you understand.

CB