# Thread: exam problems (calculus 1)

1. ## exam problems (calculus 1)

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

2. 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

$\chi$ $\sigma$

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

4. 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

$\chi$ $\sigma$

5. 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

$\chi$ $\sigma$

6. 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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