Practical:Chain rule for differentiation

This article considers practical aspects of the chain rule for differentiation: how is this rule used in actual computations?

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Statement to remember

The statement of the chain rule for differentiation that we will be using is:

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

where $f'(u)={\frac {d(f(u))}{du}}$ and $g'(x)={\frac {d(g(x))}{dx}}$.

NOTE: As a matter of convention, and to reduce confusion, we use a different variable ($u$ in this case) for the generic input to $f$ compared to the variable ($x$ in this case) that we use for the generic input to $g$.

Procedure to apply the chain rule for differentiation

The chain rule for differentiation is useful as a technique for differentiating functions that are expressed in the form of composites of simpler functions.

Most explicit procedure

The explicit procedure is outlined below:

1. Identify the two functions whose composite is the given function. In other words, explicitly decompose the function as a composite of two functions. We will here call the functions $f$ and $g$, though you may choose to give them different names.
2. Calculate the derivatives of $f$ and $g$ separately, on the side.
3. Plug into the chain rule formula the expressions for the functions and their derivatives.
4. Simplify the expression thus obtained (this is optional in general, though it may be required in some contexts).

For instance, consider the problem:

Differentiate the function $p(x):=\sin(x^{2})$

The procedure is as follows:

1. Identify the two functions: The two functions are $f(u)=\sin u$ and $g(x)=x^{2}$ (note: per the note included with the formulation of the chain rule, we use different variable names for the generic variable for the two functions, to reduce confusion regarding which one to apply on what).
2. Calculate the derivatives: $f'(u)=\cos u$ and $g'(x)=2x$.
3. Plug into the chain rule formula: We get $p'(x)=f'(g(x))g'(x)=\cos(x^{2})(2x)$.
4. Simplify the expression thus obtained: There isn't really anything to simplify, but we can rearrange the terms to the more conventional order where the algebraic part is before the trigonometric part, obtaining the final answer $p'(x)=2x\cos(x^{2})$.

More inline procedure using Leibniz notation

Although the explicit procedure above is fairly clear, Step (2) of the procedure can be a waste of time in the sense of having to do the derivative calculations separately. If you are more experienced with doing differentiation quickly, you can combine Steps (2) and (3) by calculating the derivatives while plugging into the formula, rather than doing the calculations separately prior to plugging into the formula. Further, we do not need to explicitly name the functions if we use the Leibniz notation to compute the derivatives inline.

The shorter procedure is outlined below:

1. Identify the two functions being composed (but you don't have to give them names).
2. Plug into the formula for the chain rule, using the Leibniz notation for derivatives that have not yet been computed.
3. Compute derivatives and simplify

For instance, consider the problem:

Differentiate the function $p(x):=\sin(\ln x)$

1. Identify the two functions being composed: The functions are $u\mapsto \sin u$ (the outer/later function) and $x\mapsto \ln x$ (the inner/earlier function)
2. Plug into the formula for the chain rule: We get $p'(x)={\frac {d(\sin(\ln x)}{d(\ln x)}}{\frac {d(\ln x)}{dx}}$ (here basically $u=\ln x$, though we don't have to say this explicitly)
3. Compute derivatives and simplify: We get $p'(x)=\cos(\ln x){\frac {1}{x}}={\frac {\cos(\ln x)}{x}}$

Shortest inline procedure

If you are really experienced with doing derivatives in your head, you can shorten the procedure even further by combining Steps (2) and (3) in the previous procedure. The procedure has two steps:

1. Identify the two functions being composed (but you don't have to give them names).
2. Use the formula for the chain rule, computing the derivatives of the functions while plugging them into the formula

For instance, consider the problem:

Differentiate the function $q(x):=\ln(1+\ln x)$

1. Identity the two functions: The functions are $u\mapsto \ln u$ (the outer/later function) and $x\mapsto 1+\ln x$ (the inner/earlier function)
2. Use the formula for the chain rule, computing the derivatives of the functions while plugging them into the formula: We get $q'(x)={\frac {1}{1+\ln x}}{\frac {1}{x}}={\frac {1}{x(1+\ln x)}}$