The chain rule, and differentiating a relation without first solving it for y

What to assume they know

• The derivative as a limit, and the sum, product and quotient rules
• Table 5.3: the derivatives of a whole power, the sine, the cosine and the tangent
• Composition of functions, and naming the two functions that make up a composite
• The binomial expansion of a small power of a linear expression
• Trigonometric identities used for rewriting, from Class XI
• Solving a linear equation for one unknown, where the unknown appears in several terms
• The idea that a relation between two letters may determine one of them without giving a formula for it

What they should be able to do

• Recognise a composite and name its inner and outer functions before differentiating
• State the chain rule in the chapter's own notation and apply it to a two-stage composite
• Extend the rule to a three-stage composite and identify the three factors
• Differentiate the exercise set of §5.3.1 without expanding anything
• Distinguish a relation solvable for the second letter from one that is not, and give an instance of each
• Differentiate both sides of a relation with respect to the first letter, treating the second as an unknown function
• Handle the derivative of a constant appearing on one side of a relation
• Collect the unknown derivative from several terms and solve for it
• State the inputs at which an implicit answer is undefined, from the denominator that appears
• Recognise that an answer containing both letters is a complete answer

Where it usually goes wrong

• "The chain rule was proved." It was checked on one case and then stated. The chapter's own sentence says the proof is skipped. A student should not claim to have derived it.
• "Differentiate the inside and multiply by the outside." The other way round: the derivative of the outer rule, evaluated at the inner output, times the derivative of the inner rule. Getting the evaluation point wrong is the single commonest error in this topic.
• "Leave the answer in terms of the intermediate letter." The chapter says outright that the result is normally expressed in the original letter. Substitute back every time.
• "Implicit differentiation is a different rule." It is the chain rule applied to a letter that stands for an unknown function of the other. Nothing new is assumed except that the second letter does depend on the first.
• "If the relation can be solved, implicit differentiation is wrong." Example 22 solves it both ways and gets the same answer. Being solvable is a convenience, not a constraint.
• "An answer containing the second letter is incomplete." For an implicit relation it is normally the only available form. Example 23's answer keeps both letters and the chapter presents it as finished.
• "The derivative of a constant term is that constant." It is zero. The chapter stops to spell this out in Example 22 because a constant on the right of a relation looks unlike a constant function until you say so.
• "Every implicit answer needs a condition attached." Only where a denominator can vanish. Example 23 needs one; Example 22 does not. Attach the condition when the denominator earns it, not by reflex.

Questions to check understanding

• Differentiate a two-stage composite, naming the inner and outer rules first — the form of Exercise 5.2 Q1 to Q3 and Q8
• Differentiate a three-stage composite and exhibit the three factors — the form of Exercise 5.2 Q4 and Q7
• Differentiate a composite that also needs the product or quotient rule — the form of Exercise 5.2 Q5 and Q6
• Differentiate a relation implicitly and solve for the derivative — the form of Exercise 5.3 Q1 to Q6
• Differentiate a relation in which a product of both letters sits inside a trigonometric function — the form of Exercise 5.3 Q7
• State the inputs at which an implicit answer fails, reading them off the denominator — the form of Example 23
• Prove a stated closed form for an implicit derivative using a trigonometric identity — the form of Miscellaneous Exercise Q16

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• The motivating computation (§5.3.1, Part I pp. 120–121). Take twice the input plus one, all cubed. The chapter expands it into a cubic, differentiates term by term, and then factorises the result back. Verified: the expansion is eight times the cube, plus twelve times the square, plus six times the input, plus one; differentiating gives twenty-four times the square plus twenty-four times the input plus six; and that factorises to six times the square of the original bracket. Every number here should be shown, because the whole argument of section 2 is that the last factorisation is not a coincidence.
• The re-reading as a composite (Part I p. 121). The chapter names the inner rule as twice the input plus one and the outer rule as cubing, introduces a letter for the inner output, and rewrites the answer as three times the square of that letter, times two. Verified: three times the square of the bracket times two is six times the square of the bracket — the same expression the expansion produced. The point of the example is that both routes exist here and only one of them exists for a hundredth power, which the chapter says outright in the next sentence.
• Theorem 4 (Part I p. 121). The chain rule, stated for a composite of two functions with a named intermediate letter, requiring both derivatives to exist. The chapter then says it is skipping the proof. That sentence belongs in the explanation next to the statement. Of the chapter's five numbered theorems only two carry any proof at all — the first, and only its opening part, and the third — while the second, this one and the fifth are stated and left where they stand.
• The three-function extension (Part I p. 121). A composite of three functions with two intermediate letters, differentiated as a product of three factors, provided all of them exist. The chapter invites the reader to write out the general case. Verified: the pattern is one factor per stage, each the derivative of that stage with respect to the letter feeding it.
• Example 21 (Part I pp. 121–122). The sine of a square. The chapter names the inner and outer rules, checks that both derivatives exist, multiplies, and then substitutes back so that the answer contains only the original letter. Verified: the answer is twice the input times the cosine of the square. The substitution back is a printed instruction, not an afterthought — the chapter says it is normal practice, and students who leave the intermediate letter in lose marks for no mathematical reason.
• Exercise 5.2 Q1 to Q8 (Part I p. 122). Eight items, all pure chain rule. Verified, in order: Q1 gives twice the input times the cosine of the square plus five. Q2 gives minus the sine of the sine, times the cosine. Q3 gives the first constant times the cosine of the linear expression. Q4 needs three stages — a square root inside a tangent inside a secant — and gives the secant of the tangent times its own tangent, times the secant squared of the root, all over twice the root. Q5 is a quotient with a chain rule in each part. Q6 is a product of two composites and needs the product rule as well. Q7 has a root outside a cotangent outside a square. Q8 is a cosine of a root. Q4, Q6 and Q7 are the three that combine the chain rule with something else.
• The two opening relations of §5.3.2 (Part I p. 122). One is a difference of the two letters set against a constant; the other has one letter inside a sine of the product of both. Verified: the first can be rearranged for the second letter in one step; the second cannot be rearranged at all by elementary means, and the chapter says so. The pair is the whole justification for the section and should be shown together.
• Example 22 (Part I p. 123). The solvable relation, done twice: once by rearranging first and once by differentiating the relation as it stands. Both give one. Verified. The chapter also stops to say what differentiating a constant means. Choose this example for section 8 precisely because the answer is already known — the method is being tested against a case where a mistake would be obvious.
• Example 23 (Part I p. 123). A sum of the second letter and its sine, set against the cosine of the first. Differentiating gives the derivative plus the cosine of the second letter times the derivative on the left, and minus the sine of the first on the right. Verified: collecting gives the derivative as minus the sine of the first letter over one plus the cosine of the second, and the printed condition — that the second letter avoid the odd multiples of pi — is exactly where that denominator vanishes. Check the condition; it is the only place in this topic where the chapter attaches one, and it is attached correctly.
• Exercise 5.3 Q1 to Q8 (Part I p. 125). Eight relations to differentiate. Verified, in order: Q1 gives the cosine of the first letter minus two, all over three. Q2 gives two over the cosine of the second letter minus three. Q3 gives minus the first constant over twice the second constant times the second letter plus the sine of the second letter. Q4 gives the secant squared minus the second letter, over the first letter plus twice the second minus one. Q5 gives minus twice the first plus the second, over the first plus twice the second. Q6 gives minus three times the first squared plus twice the product plus the second squared, over the first squared plus twice the product plus three times the second squared. Q7 gives the second letter times the sine of the product, over the sine of twice the second letter minus the first letter times the sine of the product. Q8 gives the sine of twice the first over the sine of twice the second. Q5 and Q6 are the two where the product rule is unavoidable, and Q7 is the one where it is needed inside a chain rule.
• Miscellaneous Exercise Q16 (Part I p. 145). The cosine of the second letter equals the first letter times the cosine of a fixed angle plus the second letter; the target is a stated closed form. Verified: differentiate both sides, substitute for the first letter from the original relation, collect the derivative over a common denominator, and the numerator that appears is the sine of a difference of two angles which collapses to the sine of the fixed angle alone. The printed answer follows. This is the hardest implicit item in the chapter and it is the only one that needs a trigonometric identity as well as the algebra; give it its own section.

Figures to have open

• A two-box series diagram for section 2, with the inner and outer derivatives attached to the boxes rather than to the arrows, so the product is visible when the second box is added. Not in the book; §5.3.1 and §5.3.2 print no figure at all — verified on the page image of every page from Part I p. 118 to Part I p. 125.
• A three-box version of the same diagram for section 4, built by extending the first rather than by drawing a new one.
• A sorted table of Exercise 5.2's eight items for section 6, with a column naming what each needs beyond the chain rule. Use the repo's DataTable component.
• A side-by-side of the two opening relations of §5.3.2 (Part I p. 122) for section 7, one shown being rearranged and one shown resisting rearrangement.
• No redraw of any textbook figure is needed anywhere in this topic, because there is none to redraw.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 5, §5.3.1 Derivatives of composite functions, the motivating computation, Part I pp. 120–121
• Theorem 4 with the skipped proof and the three-function extension, Part I p. 121; Example 21, Part I pp. 121–122
• Exercise 5.2, questions 1 to 8, Part I p. 122
• §5.3.2 Derivatives of implicit functions, the two opening relations, Part I p. 122; Examples 22 and 23, Part I p. 123
• Exercise 5.3, questions 1 to 8, Part I p. 125; Miscellaneous Exercise on Chapter 5, question 16, Part I p. 145; Summary, the chain rule bullet, Part I p. 146