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The general equation is:

$dY = {\partial Y\over\partial L}*dL+{\partial Y\over\partial K}*dK$

I know partial differentiation, and I was told, that $dL$ means total differentiation of L, but in this example L is a function argument.

This is for a Cobb-Douglas-Function. My example:

$Y(L,K) = L^{0.7}*K^{0.3}$

So you can see that L and K are arguments. How do I differentiate them?

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    "So you can see that L and K are constants." — No, I can't. To me, they look like function arguments. – 2017-02-12
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    Well, how to take a total derivative of a function argument? – 2017-02-12
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    I don't know what they want you to do. Did they ask a question? – 2017-02-12
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    "The total derivative of the CD-Function is the general isoquant equation" - "Calculate to general isoquant equation to this production function in the form of L(K)" is all I've got. But I already fail to understand what $dL$ is meant to mean. – 2017-02-12
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    @Thyzer The $dL$ is the change in $L$. It's not something you can compute (unless you're given some means to tell how much $L$ has changed in your particular problem). I'm not sure what an isoquant is, but I put an answer down about how to interpret the equation on the Cobb douglas form. Hope it helps. – 2017-02-12
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    @Thyzer A google search tells me an isoquant means $Y$ doesn't change, so the linearized isoquant equation means $0 = 0.7dL/L + 0.3dK/K$ which you can integrate to get $L^{0.7} = CK^{-0.3}.$ Of course this would also be what you get if you set $Y = K^{0.3}L^{0.7} =C$, i.e. setting $Y$ to a constant. – 2017-02-12

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You can take the partial derivatives of $Y$ and plug in to get the linearized change in $Y$ $$dY = 0.7L^{-0.3}K^{0.3}dL + 0.3L^{0.7}K^{-0.7}dK = 0.7Y\frac{dL}{L} + 0.3Y\frac{dK}{K} $$ so that $$\frac{dY}{Y} = 0.7\frac{dL}{L} + 0.3\frac{dK}{K}$$

This tells you how the percentage change in $Y$ relates to the percentage change in $L$ and $K$ (for small changes). So a $10\%$ change in $L$ without any change in $K$ causes about a $7\%$ change in $Y.$

To find your isoquants, that means $Y$ isn't changing so $dY/Y=0$ and you get $$ 0 = 0.7\frac{dL}{L} + 0.3\frac{dK}{K} = 0.7d(\log L) + 0.3d(\log K).$$

Integrating this gives $$ 0.7\log L = -0.3\log K + C$$ or $$ L= C'K^{-0.3/0.7}$$ where $C'$ is an arbitrary constant.

We could have also come to this by just setting $Y=C,$ i.e. set $Y$ equal to a constant. Then $$ C = L^{0.7}K^{0.3} \Rightarrow L = C'K^{-0.3/0.7}$$ where $C' = Y^{1/0.7}$