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Here is a simple question. Let $y=f(x)$ and $z=g(x)$ are two well behaved functions of $x$. How can I calculate the derivative $\frac{dy}{dz}$?

Related question is about 2-variable functions $y=f(x_1,x_2)$ and $z=g(x_1,x_2)$. The question is how to calculate the following derivatives: $$ \left.\dfrac{\partial y}{\partial z}\right|_{x_1=\text{const}} \;\;\;\text{and}\;\;\; \left.\dfrac{\partial y}{\partial z}\right|_{x_2=\text{const}} $$ I have a feeling that the inverse functions are involved here but I'm not really sure how to proceed.

Thanks!

Alexander S
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1 Answers1

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For the one-variable functions: $$\frac{\text d y}{\text d z} = \frac{\text d y}{\text d x} \cdot \frac{\text d x}{\text d z} = \frac{f'(x)}{g'(x)}$$

For the two-variable functions:

$$\frac{\partial y}{\partial z} = \frac{\partial y}{\partial x_1} \frac{\partial x_1}{\partial z} + \frac{\partial y}{\partial x_2}\frac{\partial x_2}{\partial z}$$

Now, depending on whether $x_1$ or $x_2$ is a constant, one of the terms becomes zero.

EDIT: Let's see what $\frac{\partial x_1}{\partial z}$ is when $x_2=\text{const}$. The same reasoning applies to $\frac{\partial x_2}{\partial z}$ when $x_1=\text{const}$.

Since $x_2$ is constant, then $z=g(x_1,x_2)$ is just a function of $x_1$. Because of this, we can define a new function $$h(x_1) = g(x_1,x_2=\text{const})$$ and we have that $z=h(x_1)$. In the case of a one-variable function, a partial derivative is the same as an ordinary derivative. Hence $$\frac{\partial x_1}{\partial z}=\frac{\partial x_1}{\partial h(x_1)} = \underbrace{ \frac{\text d x_1}{\text d h(x_1)} = \frac{1}{h'(x_1)} }_\text{inverse function rule} = \frac{1}{\dfrac{\partial z}{\partial x_1}}$$

Haris Gušić
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  • Thanks! The case with the one-variable function is indeed trivial. As to the two-variable functions case I'm still confused as to how to interpret $\partial x_i/\partial z$. Are they understood as follows? $$ \dfrac{\partial x_1}{\partial z} = \dfrac{\partial g^{-1}(z,x_2=\text{const})}{\partial z} $$ $$ \dfrac{\partial x_2}{\partial z} = \dfrac{\partial g^{-1}(z,x_1=\text{const})}{\partial z} $$ where $g^{-1}(z)$ are the inverse functions subject to constraints. – Alexander S Feb 25 '19 at 04:39