Solow model: Partial derivative of y with respect to n I am trying to find the derivative of $y$ with respect to $n$. All I need to do is take the partial derivative $\frac{\partial y}{\partial n}$ of the function: $sf(k) = (n+g+\delta)k$  Here is what I get using the chain rule, but it's wrong:

$$sf'(k)  \frac{\partial k}{\partial n} = (n+g+\delta)\frac{\partial k}{\partial n}$$

To solve $\frac{\partial y}{\partial n}$ some more steps are required, but as far as I know the equation is already wrong. From what I have seen the right side should be augmented by k and then it would be right. Any ideas where the derivation went wrong?
You need to use the implicit function theorem. The long-run relationship implies

$$
0 = F(n,k) = s f(k) - (n+g+\delta)k,
$$

and this implicitly defines function $k(n)$. The derivative is then

$$
\frac{d k(n)}{d n} = -\frac{\partial F / \partial n}{\partial F / \partial k} = \frac{k}{s f'(k) - (n+g+\delta)},
$$

which can be evaluated once one has solved for particular long-run (n,k) combination. To finally obtain derivative of output $y = f(k)$ wrt. $n$, simply use chain rule:

$$
\frac{d y}{d n} = f'(k(n)) \frac{d k(n)}{d n}.
$$