Implicit Differentiation Help!

Question

Could use some help with this -

Given $e^L + KL = Ke^K$, we are being asked to find $\frac{dL}{dK}$. I think O need to use implicit differentiation, but I am really not sure how to do it!

Answer 1

Consider the implicit function $$f(K,L)=e^L + KL - Ke^K=0$$ Now $$\frac{\partial f(K,L)}{\partial K}=L-e^K (K+1)\qquad \text{and} \qquad \frac{\partial f(K,L)}{\partial L}=e^L+K$$ Now, using the implicit function theorem $$\frac{dL}{dK}=-\frac{\frac{\partial f(K,L)}{\partial K} } { \frac{\partial f(K,L)}{\partial L}}=\frac{e^K (K+1)-L}{K+e^L}$$

Answer 2

There are a lot of different ways to do implicit differentiation. Some methods have you differentiate respect to a particular variable, but, the way I prefer is to just do differentiation without being with respect to a particular variable, and then solve for the derivative you are looking for. So, instead of the derivative rule for $y = u^n$ being $\frac{dy}{du} = nu^{n-1}$, I instead have the differential rule $dy = nu^{n-1}\, du$. Then I can algebraically divide by $du$ if I want to find $\frac{dy}{du}$.

So, in your case, we have: $$ e^L + KL = Ke^K $$ Differentiating both sides gives us: $$ d\left(e^L + KL\right) = d\left(Ke^K\right) \\ d\left(e^L\right) + d\left(KL\right) = d\left(Ke^K\right) \\ e^L\,dL + K\,dL + L\,dK = Ke^K\,dK + e^K\,dK $$

now we just need to gather our terms together so we can solve for $\frac{dL}{dK}$: $$ e^L\,dL + K\,dL + L\,dK = Ke^K\,dK + e^K\,dK \\ e^L\,dL + K\,dL = Ke^K\,dK + e^K\,dK - L\,dK \\ \left(e^L + K\right)\,dL = \left(Ke^K + e^K - L\right)\,dK \\ \frac{dL}{dK} = \frac{Ke^K + e^K - L}{e^L + K} $$ More information about this method of implicit differentiation can be found here.