**#1.** Find the derivative of: $\displaystyle e^x^y=sin(y^2)$

$\displaystyle lne^x^y=lnsin(y^2)$

$\displaystyle xylne=lnsin(y^2)$

$\displaystyle xy*1=lnsin(y^2)$

$\displaystyle y+xy'=lnsin(y^2)$

I need to use 2 chain rules on $\displaystyle lnsin(y^2)$

$\displaystyle y+xy'=[{\frac{1}{sin(y^2)}]*[cos(y^2)*(2yy')]$

$\displaystyle y+xy'=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}$

$\displaystyle y=\frac{[cos(y^2)*(2yy')]}{\frac{1}{sin(y^2)}}}-xy'$

$\displaystyle y=y'[\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}}-x]$

$\displaystyle y'=\frac{y}{\frac{[cos(y^2)*(2y)]}{\frac{1}{sin(y^2)}}-x}}$

I'm sure there is something wrong in there

**#2.** Find the derivative using logartihmic differentiation:

$\displaystyle y=\frac{x(x+1)^3}{(3x+1)^2}$

$\displaystyle lny=ln\frac{x(x+1)^3}{(3x+1)^2}$

Now I am unsure of what to do. Can I, and if so should I, rewrite the right side as:

$\displaystyle lny=[ln(x(x+1)^3]-[ln(3x+1)^2]$

Or should I use a combination of the quotient, product and multiple chain rules on what I had before:

$\displaystyle lny=ln\frac{x(x+1)^3}{(3x+1)^2}$

You don't have to show me any work past this point, I would like to figure this our on my own, I just need a push in the right direction.