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Given a multivariate composition like ''f(g,h)'' where ''g = g(x,y)'' and ''h = h(x,y)'', and using a notation that expresses [[Calculus/PartialDerivative|partial derivatives]] as ''a,,b,, = ∂a/∂b'', the [[Calculus/Differential|total differential]] demonstrates that: Equivalently, consider:

{{attachment:notation1.svg}}

{{attachment:notation2.svg}}



=== Multivariate ===

A notation for [[Calculus/PartialDerivative|partial derivatives]] will be used such that ''a,,b,, = ∂a/∂b''.

Given a multivariate composition like ''f(g,h)'' where ''g = g(x,y)'' and ''h = h(x,y)'', the [[Calculus/Differential|total differential]] demonstrates that:
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Clearly then ''f,,g,,g,,x,, + f,,h,,h,,x,, = f,,x,,'' and ''f,,g,,g,,y,, + f,,h,,h,,y,, = f,,y,,'', so: Clearly then ''f,,g,,g,,x,, + f,,h,,h,,x,, = f,,x,,'' and ''f,,g,,g,,y,, + f,,h,,h,,y,, = f,,y,,''. Equivalently, consider:

{{attachment:n
otation3.svg}}

Chain Rule

The chain rule is an approach for differentiating a composition of functions.


Description

Given a function defined like f(x) = g(h(x)), the derivative of f is g'(h(x)) h'(x).

Sometimes this is notated as f = g ∘ h and f' = (g' ∘ h)h' .

Equivalently, consider:

notation1.svg

notation2.svg

Multivariate

A notation for partial derivatives will be used such that ab = ∂a/∂b.

Given a multivariate composition like f(g,h) where g = g(x,y) and h = h(x,y), the total differential demonstrates that:

df = fgdg + fhdh

df = fg(gxdx + gydy) + fh(hxdx + hydy)

df = fggxdx + fggydy + fhhxdx + fhhydy

df = (fggx + fhhx)dx + (fggy + fhhy)dy

Clearly then fggx + fhhx = fx and fggy + fhhy = fy. Equivalently, consider:

notation3.svg


Usage

Point elasticity is defined as:

elasticity1.svg

It can be simplified using the chain rule into:

elasticity2.svg

Start with the final formulation and use u- and v-substitution:

substitution1.svg

Note in particular that the equation for v can be rewritten in terms of P:

substitution2.svg

Now the final formulation can be rewritten as:

substitution3.svg

Several of these terms are trivial to derive. The derivative of e to an exponent is itself, so the derivative of P = ev with respect to v is P. The derivative of a natural logarithm of some variable is 1 over that variable, so the derivative of u = ln Q with respect to Q is 1/Q. This leaves the equation as:

substitution4.svg


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Calculus/ChainRule (last edited 2025-10-16 20:27:40 by DominicRicottone)