calculus

sensitivity analysis

rules

chain rule

$\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x)$

product rule

$\frac{d}{dx} (f(x) g(x)) = f(x) g'(x) + g(x) f'(x)$

derivative with respect to a function

$\frac{d}{df} y$

I am not clear on how to handle this situation.

I hypothesize that the result depends on whether or not $f$ is a function of $y$ . And that if $f$ is a function of $y$ , it implies that $y$ is a function of $f$ .

We know that if $f$ is not a function of $y$

$\frac{df}{dy} = 0$

It may be helpful to stop thinking about variables as being different that functions of variables.

I can show for some examples that

$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$

where

$y = f(x)$

Here are the examples

$y = ax$ $\frac{dy}{dx} = a$ $x = \frac{1}{a} y$ $\frac{dx}{dy} = \frac{1}{a}$

And another

$y = e^x$ $\frac{dy}{dx} = e^x$ $x = \ln(y)$ $\frac{dx}{dy} = \frac{1}{y} = \frac{1}{e^x}$

What if $x$ and $y$ are inseparable?

$y = x + e^x$ $dy = dx + e^x dx$ $\frac{dy}{dx} = 1 + e^x$ $\frac{dx}{dy} = \frac{1}{1 + e^x}$

Charles Rymal

My Blog

calculus

rules

chain rule

product rule

derivative with respect to a function