# Thread: a derivative and 1 limit

1. ## a derivative and 1 limit

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

2. 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.