a derivative and 1 limit

**wanqwer** · March 2nd 2009, 03:09 PM

Hello everyone,
i'm having my exam tomorrow, now doing homework and trying to solve some problems. Here are two that are too hard for me

. If anyone could help me solve it...
Any help will be appreciated

1 is to find derivative. $\cos\sqrt{x^2+y^2}=\sin^2x+(\frac{1}{y})$
i've done this:
$\cos\sqrt{x^2+y^2}-\sin^2x-(\frac{1}{y})=0$
then trying to find derivatives of each part

part 1 $(\cos\sqrt{x^2+y^2})'=-\sin\sqrt{x^2+y^2}*(\sqrt{x^2+y^2})'=$
$=-\sin\sqrt{x^2+y^2}*\frac{1}{2\sqrt{x^2+y^2}}*(x^2+ y^2)'=$
$=-\frac{\sin\sqrt{x^2+y^2}}{2\sqrt{x^2+y^2}}*(2x*2y)$
part 2 $(-\sin^2x)'=$ can't remember this

part 3 $(\frac{1}{y})'=-\frac{1}{y^2}$

2 limit. $\lim_{x\to\infty}(\frac{3x-1}{2x+3})^x$ not even know how to start

**matheagle** · March 2nd 2009, 03:54 PM

Originally Posted by wanqwer

Hello everyone,
i'm having my exam tomorrow, now doing homework and trying to solve some problems. Here are two that are too hard for me

. If anyone could help me solve it...
Any help will be appreciated

1 is to find derivative. $\cos\sqrt{x^2+y^2}=\sin^2x+(\frac{1}{y})$
i've done this:
$\cos\sqrt{x^2+y^2}-\sin^2x-(\frac{1}{y})=0$
then trying to find derivatives of each part

part 1 $(\cos\sqrt{x^2+y^2})'=-\sin\sqrt{x^2+y^2}*(\sqrt{x^2+y^2})'=$
$=-\sin\sqrt{x^2+y^2}*\frac{1}{2\sqrt{x^2+y^2}}*(x^2+ y^2)'=$
$=-\frac{\sin\sqrt{x^2+y^2}}{2\sqrt{x^2+y^2}}*(2x*2yy ')$
part 2 $(-\sin^2x)'=$ can't remember this

part 3 $(\frac{1}{y})'=-\frac{1}{y^2}$

2 limit. $\lim_{x\to\infty}(\frac{3x-1}{2x+3})^x$ not even know how to start

The first problem is your notation.
By ' I'm assuming you mean ${d\over dx}$
So when you differentiate x wrt s you get one, but when you differentiate
y wrt x you get ${dy\over dx}$ and not 1.
NOTE.... I inserted that y' just now in my copy of your solution to 1

$(-\sin^2x)'=-2(\sin x)^1(\cos x)=-\sin (2x)$

$(\frac{1}{y})'=-\frac{1}{y^2}{dy\over dx}$ UNLESS the prime means ${d\over dy}$ .

For $\lim_{x\to\infty}(\frac{3x-1}{2x+3})^x$
you can divide inside the () and get (1+...).
BUt it's better to learn how to do this by scratch.
Let $y=(\frac{3x-1}{2x+3})^x$ , then take log's of both sides, that pulls down the exponent x.
Work on this and I'll check it later.
The point to this problem is to make it into a fraction and apply L'Hopitals' Rule.

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