Applying the Chain Rule to a Composite Multivariable Function

Summary

The document asks whether the derivative of a composite function y=f(x), with x depending on a and b, equals the sum of the partial derivatives of y with respect to those variables. It tests that proposed relationship with a simple counterexample: let y equal x and let x equal the sum of a and b. The derivative of y with respect to x is then one, while the two partial derivatives with respect to a and b sum to two.

This shows that the stated formula is generally incorrect and illustrates why derivatives with respect to different variables cannot simply be added as proposed. The document does not provide the corrected multivariable chain rule or discuss applications in finance, so its instructional scope is limited to identifying the error with one example.

Key ideas

• A function that depends on intermediate variables requires the chain rule to relate its derivatives.
• The derivative with respect to x is not generally the sum of partial derivatives with respect to a and b.
• A simple linear example demonstrates that the proposed equality can fail.

Tags

# partial derivatives of multivariable function


# partial derivatives of multivariable function












Looking to verify whether the following formulation is correct. Suppose we have the following function, relationships:

$$y=f(x)$$ $$x=g(a,b)$$ $$y=f[g(a,b)]$$

Is the below correct (including notation)? $$\frac{dy}{dx}=\frac{\partial y}{\partial a}+\frac{\partial y}{\partial b}$$ $$\frac{dy}{dx}=\frac{\partial y}{\partial x}\frac{\partial x}{\partial a}+\frac{\partial y}{\partial x}\frac{\partial x}{\partial b}$$

In words, the total derivative of the composite function $y$ with respect to $x$ is the sum of the partial derivatives of $y$ with respect to $a$ and $b$

## Answer by amars (score 2)

https://quant.stackexchange.com/a/50025

No, this is in general not true. For example, consider \begin{align*} y(x)&:=x,\\ x(a,b):&=a+b. \end{align*} Then we have $$\frac{\partial y}{\partial x}=1$$ but $$\frac{\partial y}{\partial a} + \frac{\partial y}{\partial b}=2. $$

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