Derivative of a function where $y$ is implicit function of $x$ So I have an equation that determines y implicitly as function of $x$. My job is to find $y''(0)$
The equation is
$$
 \ln(1+y)+\sin(xy)=\ln(5)
$$
What I've tried doing is just raising everything as a power of eulers number, so i can cancel the ln terms.
Therefore I'm left with
$$
 1+y+e^{\sin(xy)}=5 
$$
which then gives that
$$
y=4-e^{\sin(xy)}
$$
I then take and differentiate twice and input $y''(0)$ which leaves me with $e^0=1$. It seems wrong, what have i done wrong here?
 A: I see no reason to try to solve for y.  Implicit differentiation works nicely.
The original equation is $\ln(1+ y)+ \sin(xy)= \ln(5)$.   The derivative of y, with respect to x, is given by $\frac{y'}{1+ y}+ \cos(xy)(y+ xy')= 0$.
Note that when x= 0, $\ln(1+ y(0))+ \sin(0)= ln(5)$ so that $1+ y(0)= 5$ and $y(0)= 4$.  then $\frac{y'(0)}{5}+ \cos(0)(y(0)+ 0)= \frac{y'(0)}{5}+ 4= 0$.  So $y'(0)= -20$.
Using implicit differentiation again, $\frac{y''}{1+ y}- \frac{y'^2}{(1+ y)^2}- \sin(xy)(y+ xy')^2+ \cos(xy)(1+ y'+ xy'')= 0$.   Setting x= 0, $\frac{y''(0)}{1+ y(0)}- \frac{y'(0)^2}{(1+ y(0))^2}- \sin(0y(0))(y(0)+ 0)^2+ cos(0y(0))(1+ y'(0)+ 0y''(0))= \frac{y''(0)}{5}- \frac{400}{25}- 19= \frac{y''(0)}{5}- 3= 0$ so $y''(0)= 15$.
A: Raising everything as a power of $e$ yields
$$e^{\ln(1+y)+\sin(xy)}=e^{\ln 5}\\
(1+y)\cdot e^{\sin(xy)}=5$$
A: After composing with the exponential function, perform some elementary differential calculus: from $\:(1+y)\mathrm e^{\sin(xy)}=5$, you deduce instantly that
\begin{align}
\mathrm d\bigl((1+y)\,\mathrm e^{\sin(xy)}\bigr)&=\bigl((1+y)\,\mathrm e^{\sin(xy)}y\cos(xy)\bigr)\mathrm dx\\
&\phantom{=}+\bigl(\mathrm e^{\sin(xy)}+(1+y)\mathrm e^{\sin(xy)}x\cos(xy)\bigr)\mathrm dy.
\end{align}
Can you deduce $\;\dfrac{\mathrm dy}{\mathrm dx}\:$ from there?
