Derivate of a complex function

Question

I found in a paper the folowing derivative:

$$\frac{\partial f(X(\theta),\theta)}{\partial \theta}=\frac{\partial f(X(\theta),\theta)}{\partial X(\theta)}\frac{\partial X(\theta)}{\partial \theta}+\frac{\partial f(X(\theta),\theta)}{\partial \theta}$$

How they get this derivative?

Answer 1

This is bad notation. It could instead be written as

$$g(\theta)=f(X(\theta),\theta)$$

$$g'(\theta)=f^{(1,0)}(X(\theta),\theta)\cdot X'(\theta)+f^{(0,1)}(X(\theta),\theta)$$

where $f^{(p,q)}$ denotes a partial derivative of $f$. This is the multivariable chain rule.

See also this.

Answer 2

You combine the way you differentiate a function of two variables with the chain rule.

I find it best to explain this with differentials (others may disagree, and provide another answer with a rigorous proof).

If you have a function $f(x,y)$ of two variables and want to know how small changes in $x$ and $y$ cause $f$ to change you calculate

$$ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy $$

Then if each of $x$ and $y$ depends on some parameter $t$ you know $$ \frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} . $$

You can see how your (confusingly written) expression matches this pattern. There the dependence of the second variable on the parameter is simply $y = t$ so $dy/dt = 1$.