exam problems (calculus 1)

**eq123** · December 14th 2010, 07:52 AM

hey everyone, i would appreciate any help with these problems ..

**chisigma** · December 14th 2010, 08:09 AM

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)

Merry Christmas from Italy

$\chi$ $\sigma$

**Mondreus** · December 14th 2010, 08:16 AM

12) Rewrite it as $\ln\left[F(x)\right] = \ln\frac{\left[f(x) \right]^\pi}{\left[3+f(x)\right]^e}=\pi \ln\left[f(x) \right]-e \ln\left[3+f(x)\right]$

**chisigma** · December 14th 2010, 08:20 AM

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)

Merry Christmas from Italy

$\chi$ $\sigma$

**chisigma** · December 15th 2010, 12:17 AM

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 $F^{'} (0)$ ...

Merry Christmas from Italy

$\chi$ $\sigma$

**CaptainBlack** · December 15th 2010, 02:23 AM

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

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