Derivative of a function with respect to another function Can somebody explain what we are really doing when we are differentiating a function with respect to another function?(Not the formula just an intuitive explanation)
For derivatives with respect to x(or any other variable), I used the idea of a slope but this idea doesn't really make sense when talking about functional derivatives. 
Can somebody help me out?
 A: I can try to provide some intuition for it, confident that more competent fellow Mathstackexchangers will sort your issue out.
The expression "to differentiate with respect to another function" might be misleading. Moreover, I do not believe this is "functional derivation", which relates to functionals, a map from functions to scalars, not functions.
What is being done really, is to differentiate with respect to a transformed variable, very similar to variables change for integration.
Let me clarify with an example, $f(x) = x^2$, $f'(x) = 2x$.
The derivative with respect to a transformed variable $w(x)$ is given by the chain rule
$$ \frac{\mathrm{d}f}{\mathrm{d}w} = \frac{\mathrm{d}f}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}w}= \frac{f'}{w'}$$
A simple yet revealing case is maybe $w(x) = x^2$, so that $f = w$.
The formula above tells us it is $1$. It sounds reasonable, like the derivative of $z(x) = x$ $$\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{\mathrm{d}x}{\mathrm{d}x}.$$ 
This can be interpreted by taking the graph of $f(x) = x^2$, and transforming the horizontal axis according to the transformation $w(x)$.
One then gets a straight line, of unitary slope, in accordance with the value of the derivative given by the formula above, or the derivative of $z$.
One more example, differentiate $f(x) = x^3$ with respect to $w(x)=x^2$.
The formula above yields $ \frac{3}{2}x$.
Let us check directly by transforming the $x$-axis to the new variable $x' = x^2$, and hence $x^3 = x' \cdot \sqrt{x'} $, now we derive with respect to $x'$ and obtain
$$ \frac{\mathrm{d}f}{\mathrm{d}x'} = \frac{3}{2} \sqrt{x'},$$ and once we transform back from $x'$ to $x$ we confirm the result given by the formula.
Hope it gives some hints. One can perfectly think about it in terms of slopes, once the correct "stretching" of the $x$-axis is made.
