# Given e^L+KL=Ke^K, we are being asked to find (dL)/(dK). I think O need to use implicit differentiation, but I am really not sure how to do it!

Given ${e}^{L}+KL=K{e}^{K}$, 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!
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Lichtpulsax
Consider the implicit function
$f\left(K,L\right)={e}^{L}+KL-K{e}^{K}=0$
Now
$\frac{\mathrm{\partial }f\left(K,L\right)}{\mathrm{\partial }K}=L-{e}^{K}\left(K+1\right)\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\frac{\mathrm{\partial }f\left(K,L\right)}{\mathrm{\partial }L}={e}^{L}+K$
Now, using the implicit function theorem
$\frac{dL}{dK}=-\frac{\frac{\mathrm{\partial }f\left(K,L\right)}{\mathrm{\partial }K}}{\frac{\mathrm{\partial }f\left(K,L\right)}{\mathrm{\partial }L}}=\frac{{e}^{K}\left(K+1\right)-L}{K+{e}^{L}}$
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emmostatwf
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 $\frac{dy}{du}=n{u}^{n-1}$, I instead have the differential rule $dy=n{u}^{n-1}\phantom{\rule{thinmathspace}{0ex}}du$. Then I can algebraically divide by $du$ if I want to find $\frac{dy}{du}$.

So, in your case, we have:
${e}^{L}+KL=K{e}^{K}$
Differentiating both sides gives us:
$d\left({e}^{L}+KL\right)=d\left(K{e}^{K}\right)\phantom{\rule{0ex}{0ex}}d\left({e}^{L}\right)+d\left(KL\right)=d\left(K{e}^{K}\right)\phantom{\rule{0ex}{0ex}}{e}^{L}\phantom{\rule{thinmathspace}{0ex}}dL+K\phantom{\rule{thinmathspace}{0ex}}dL+L\phantom{\rule{thinmathspace}{0ex}}dK=K{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK+{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK$
now we just need to gather our terms together so we can solve for $\frac{dL}{dK}$:
${e}^{L}\phantom{\rule{thinmathspace}{0ex}}dL+K\phantom{\rule{thinmathspace}{0ex}}dL+L\phantom{\rule{thinmathspace}{0ex}}dK=K{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK+{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK\phantom{\rule{0ex}{0ex}}{e}^{L}\phantom{\rule{thinmathspace}{0ex}}dL+K\phantom{\rule{thinmathspace}{0ex}}dL=K{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK+{e}^{K}\phantom{\rule{thinmathspace}{0ex}}dK-L\phantom{\rule{thinmathspace}{0ex}}dK\phantom{\rule{0ex}{0ex}}\left({e}^{L}+K\right)\phantom{\rule{thinmathspace}{0ex}}dL=\left(K{e}^{K}+{e}^{K}-L\right)\phantom{\rule{thinmathspace}{0ex}}dK\phantom{\rule{0ex}{0ex}}\frac{dL}{dK}=\frac{K{e}^{K}+{e}^{K}-L}{{e}^{L}+K}$