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. $\displaystyle \cos\sqrt{x^2+y^2}=\sin^2x+(\frac{1}{y})$

i've done this:

$\displaystyle \cos\sqrt{x^2+y^2}-\sin^2x-(\frac{1}{y})=0$

then trying to find derivatives of each part

part 1 $\displaystyle (\cos\sqrt{x^2+y^2})'=-\sin\sqrt{x^2+y^2}*(\sqrt{x^2+y^2})'=$

$\displaystyle =-\sin\sqrt{x^2+y^2}*\frac{1}{2\sqrt{x^2+y^2}}*(x^2+ y^2)'=$

$\displaystyle =-\frac{\sin\sqrt{x^2+y^2}}{2\sqrt{x^2+y^2}}*(2x*2y)$

part 2 $\displaystyle (-\sin^2x)'=$ can't remember this

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

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