Practical:Chain rule for differentiation: Difference between revisions

Revision as of 05:47, 5 December 2023

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:

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.
Calculate the derivatives of $f$ and $g$ separately, on the side.
Plug into the chain rule formula the expressions for the functions and their derivatives.
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:

Identify the two functions: The two functions are $f=u\mapsto \sin u$ and $g=x\mapsto x^{2}$ (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).
Calculate the derivatives: $f'=u\mapsto \cos u$ and $g'=x\mapsto 2x$ .
Plug into the chain rule formula: We get $p'(x)=f'(g(x))g'(x)=\cos(x^{2})(2x)$ .
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 $2x\cos(x^{2})$ .

@@ Line 27: / Line 27: @@
 # '''Plug into the chain rule formula''' the expressions for the functions and their derivatives.
 # '''Simplify the expression thus obtained''' (this is optional in general, though it may be required in some contexts).
+For instance, consider the problem:
+{{quotation|Differentiate the function <math>p(x) := \sin(x^2)</math>}}
+The procedure is as follows:
+# '''Identify the two functions''': The two functions are <math>f = u \mapsto \sin u</math> and <math>g = x \mapsto x^2</math> (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).
+# '''Calculate the derivatives''': <math>f' = u \mapsto \cos u</math> and <math>g' = x \mapsto 2x</math>.
+# '''Plug into the chain rule formula''': We get <math>p'(x) = f'(g(x))g'(x) = \cos(x^2)(2x)</math>.
+# '''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 <math>2x \cos (x^2)</math>.