Chapter 2: Differentiation
Section 2.4: The Chain Rule
Example 2.4.1
Use the Chain rule to differentiate F⁡x=x4+3⁢x2+12.
Solution
Tools≻Load Package: Student Calculus 1
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Control-drag Fx=… Context Panel: Assign Function
F⁡x=x4+3⁢x2+12→assign as functionF
The derivative, simplified
F′x = 2⁢x4+3⁢x2+1⁢4⁢x3+6⁢x = 4 x 2 x2+3 x4+3 x2+1
Annotated stepwise solution
Expression palette: Differentiation template Apply to Fx.
Context Panel: Student Calculus1≻All Solution Steps (See Figure 2.4.1(a), below.)
ⅆⅆ x Fx→show solution stepsDifferentiation Stepsⅆⅆxx4+3⁢x2+12▫1. Apply the chain rule to the term x4+3⁢x2+12◦Recall the definition of the chain ruleⅆⅆxf⁡g⁡x=f'⁡g⁡x⁢ⅆⅆxg⁡x◦Outside functionf⁡v=v2◦Inside functiong⁡x=x4+3⁢x2+1◦Derivative of outside functionⅆⅆvf⁡v=2⁢v◦Apply compositionf'⁡g⁡x=2⁢x4+6⁢x2+2◦Derivative of inside functionⅆⅆxg⁡x=4⁢x3+6⁢x◦Put it all togetherⅆⅆxf⁡g⁡x⁢ⅆⅆxg⁡x=2⁢x4+6⁢x2+2⋅4⁢x3+6⁢xThis gives:2⁢x4+6⁢x2+2⁢4⁢x3+6⁢x
This calculation can be implemented interactively in the tutor, or more immediately, via the Context Panel, as shown in Figure 2.4.1(a).
Figure 2.4.1(a) Context Panel access to Differentiation Rules
Recall, we can write the composition as Fx=fgx, where fz=z2 and z=gx=x4+3 x2+1. Then, F′x is computed by the Chain rule as follows.
F′x
=ddxfgx
=ddxfzx
=ddzfzx=a|f(x)z=gx⋅dzdx
=(ddzfzx=a|f(x)z=gx)⋅g′x
In actual practice, this application of the Chain rule would most likely be implemented as follows.
=f′gx g′x
=2 gx g′x
=2x4+3 x2+1 4 x3+6 x
=4 x 2 x2+3 x4+3 x2+1
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