Getting the derivatives of the inverse trigonometric functions

What to assume they know

• The six inverse trigonometric functions with their principal-value branches, from Chapter 2 of this book
• The identity relating the square of a sine to the square of a cosine
• Implicit differentiation, and the permission to differentiate the unknown function
• The chain rule, and naming an inner and an outer function
• Where the cosine is positive and where it is negative, by quadrant
• Reading a domain written as an open interval
• Substituting a trigonometric expression for the variable to simplify an argument

What they should be able to do

• Turn an inverse trigonometric equation into a relation between the two letters and differentiate it
• Isolate the derivative and say why the answer initially contains the inverse function itself
• Identify the inputs at which the isolated derivative is undefined, and convert them into a domain
• Rewrite the cosine of an inverse sine as a square root, and justify the sign from the principal-value branch
• State the three derivatives the chapter tabulates, with their domains
• Say which of the six inverse functions get a derivative here and which do not
• Reduce a composite inverse expression by substitution before differentiating it
• Recognise a substitution that turns an expression into a constant multiple of a single inverse function
• Carry an inverse trigonometric derivative into a second-order relation

Where it usually goes wrong

• "The table is a list to memorise." Two of its three columns are derived nowhere in the chapter, and the derivations are three lines each. A student who can produce them will never mis-sign the inverse cosine, and mis-signing the inverse cosine is the commonest error in the whole chapter.
• "The inverse sine derivative works at minus one and one." It does not. The function is defined there and the derivative is not, which the chapter states explicitly on Part I p. 124.
• "The square root could be either sign." Only on the principal branch is the cosine positive, and that is what settles it. Drop the branch and the argument collapses.
• "All six inverse functions have derivatives in this chapter." Three do. The other three appear only inside exercise expressions, and never with a derivative attached.
• "An exercise using the inverse secant needs the inverse secant derivative." Exercise 5.3 Q15 does not: substitution turns it into an inverse cosine. The same for the inverse cotangent in Miscellaneous Exercise Q6.
• "The intervals attached to the exercise items are just domains." They are the conditions under which the intended collapse is valid. Ignore the interval and the substitution can produce the wrong multiple of the angle.
• "Differentiating term by term is wrong when the sum is constant." It is perfectly correct and the two terms cancel. Miscellaneous Exercise Q13 works both ways; only one of them is quick.
• "The chapter proved these functions are continuous." It stated it and said it would not prove it, and then differentiated them — which, by the chapter's own Theorem 3, is the stronger claim.

Questions to check understanding

• Derive the derivative of a named inverse trigonometric function from its defining relation, including the sign argument
• State the domain on which a tabulated inverse derivative is valid and say why the endpoints are lost
• Differentiate an inverse function of a rational expression by substituting first — the form of Exercise 5.3 Q9 to Q14
• Differentiate an expression written with an inverse function the chapter does not tabulate, by reducing it — the form of Exercise 5.3 Q15
• Differentiate a quotient one of whose factors is an inverse trigonometric function — the form of Miscellaneous Exercise Q5
• Recognise a sum of two inverse functions that is constant, and give its derivative — the form of Miscellaneous Exercise Q13
• Verify a second-order relation satisfied by an exponential of an inverse cosine — the form of Miscellaneous Exercise Q22

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 opening sentence of §5.3.3 (Part I p. 124). It asserts that the inverse trigonometric functions are continuous and says plainly that no proof will be offered. This is the fifth explicit refusal in the chapter and the pattern is worth naming once: the section is about to differentiate functions it has not shown to be continuous, having proved on Part I p. 120 that differentiability is the stronger property.
• Example 24, first half (Part I p. 124). Set the output equal to the inverse sine of the input, read it backwards as the input being the sine of the output, differentiate both sides with respect to the input, and isolate. Verified: differentiating the right-hand side needs the chain rule and produces the cosine of the output times the derivative being hunted; isolating gives one over that cosine. The answer at this stage still contains the inverse sine inside a cosine, which is the honest intermediate form and the chapter says as much before improving it.
• The domain step (Part I p. 124). The isolated answer fails where the cosine of the output vanishes; the chapter turns that into a condition on the output, then on the input, ending at the open interval from minus one to one. Verified: on the principal branch the cosine vanishes only at the two ends of the half-turn, which correspond to the inputs minus one and one, so exactly those two are lost. Notice that the domain of the derivative is strictly smaller than the domain of the function — the inverse sine itself is perfectly well defined at minus one and one.
• The square-root step (Part I p. 124). The chapter squares the identity, uses the cancellation rule for the sine of an inverse sine on the open interval, and gets the square of the cosine as one minus the square of the input. Verified. It then says that the output lies in the open half-turn from minus a quarter turn to a quarter turn, so the cosine is positive, and takes the positive root. That sentence is section 6 and it is the only place in this topic where Chapter 2's branch choice does visible work.
• The table on Part I p. 124. Three columns: the inverse sine with derivative one over the root of one minus the square, the inverse cosine with the negative of the same, and the inverse tangent with one over one plus the square; the domains are the open interval from minus one to one, the same interval, and the whole real line. Verified against the chapter's own derivation for the first column only. Three printed defects in this table are recorded in Notes, and one of them — a missing space in the last row's label — will be visible to any student.
• The two undated entries (Part I p. 124). Verified by doing the work the chapter does not: the inverse cosine derivation is Example 24 with the sine replaced by the cosine, and the sign flips because the derivative of the cosine carries a minus; the branch is the closed half-turn from zero to a half-turn, on which the sine is positive, so the root is again taken positive and the minus survives into the answer. The inverse tangent derivation replaces the sine by the tangent, whose derivative is the square of the secant, and the identity relating the square of the secant to the square of the tangent turns that into one plus the square of the input — with no sign decision at all, which is why the inverse tangent's domain is the whole line. Run at least the tangent one; it is three lines and it explains the odd column out.
• Exercise 5.3 Q9 to Q15 (Part I p. 125). Seven items, every one an inverse function of a rational or surd expression, each with an interval attached. Verified, in order: Q9 collapses to twice the inverse tangent and gives two over one plus the square. Q10 collapses to three times the inverse tangent and gives three over one plus the square. Q11 collapses to twice the inverse tangent and gives two over one plus the square. Q12 collapses to a quarter turn minus twice the inverse tangent and gives minus two over one plus the square. Q13 collapses to a quarter turn minus twice the inverse tangent and gives minus two over one plus the square. Q14 collapses to twice the inverse sine and gives two over the root of one minus the square. Q15 collapses to twice the inverse cosine and gives minus two over the root of one minus the square. The attached intervals are not decoration — each is exactly the range on which the collapse in that item is valid, and outside it the collapse is false.
• Exercise 5.3 Q15 in particular (Part I p. 125). It is written with an inverse secant, and the chapter tabulates no derivative for the inverse secant. Verified that it is nonetheless solvable with what the chapter supplies: substituting the cosine of an angle for the input turns the bracket into the secant of twice that angle, so the item is twice the inverse cosine and the tabulated inverse cosine derivative finishes it. The same is true of Miscellaneous Exercise Q6 on Part I p. 144, which is written with an inverse cotangent and collapses to half the input. Say this plainly in the explanation: no exercise in the chapter strictly needs a derivative the chapter does not print, but a student who fails to spot the substitution has nothing to fall back on.
• Miscellaneous Exercise Q4 and Q5 (Part I p. 144). Q4 is an inverse sine of the input times its own square root, on the closed unit interval. Q5 is an inverse cosine of half the input, divided by the square root of a linear expression, on the open interval from minus two to two. Verified: Q4 is the inverse sine of the three-halves power, so the chain rule gives three halves times the square root of the input, over the square root of one minus the cube; note it fails at the upper endpoint. Q5 needs the quotient rule as well as the tabulated inverse cosine derivative, and the first term of the answer carries the product of two square roots in its denominator.
• Miscellaneous Exercise Q13 (Part I p. 145). An inverse sine of the input added to an inverse sine of the root of one minus its square, on the open unit interval. Verified: the derivative is zero, because on that interval the second term is the inverse cosine of the input and the two add to a constant quarter turn. This is the best single item in the chapter for this topic: differentiating it term by term is legal and gives two expressions that cancel, and recognising the constant first gives the answer in one line.
• Miscellaneous Exercise Q22 (Part I p. 145). An exponential whose exponent is a constant times the inverse cosine, on the closed unit interval, to be shown to satisfy a second-order relation. Verified: differentiate once and multiply through by the square root of one minus the square to clear the denominator; square both sides; differentiate again and divide by twice the first derivative. The stated relation drops out. The second-order half of this item belongs to the last topic of the chapter, which carries it too; here it is the closing demonstration that these derivatives feed straight into everything after them.

Figures to have open

• A number-line drawing of the principal branch for section 4 and section 6, with the region where the cosine is positive shaded and the two removed endpoints marked. The chapter prints no figure in §5.3.3 — verified on the page image of every page from Part I p. 118 to Part I p. 125 — so this is added here.
• A three-column table for section 7 built from the table on Part I p. 124, laid out with a proper prime on the derivative row and a space in the domain label. Use the repo's DataTable component. Do not reproduce the printed labels; see Notes.
• A six-row comparison for section 9 listing the inverse functions against whether this chapter supplies a derivative. Not in the book.
• A stacked pair of substitution diagrams for section 10, showing the same expression before and after the trigonometric substitution.
• No redraw of any textbook figure is needed in this topic, because §5.3.3 contains none.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 5, §5.3.3 Derivatives of inverse trigonometric functions, the opening continuity claim and Example 24, Part I p. 124
• The unnumbered table of three inverse derivatives with their domains, Part I p. 124
• Exercise 5.3, questions 9 to 15, Part I p. 125
• Miscellaneous Exercise on Chapter 5, questions 4, 5 and 6, Part I p. 144, and questions 13 and 22, Part I p. 145
• Summary, the three inverse derivatives, Part I p. 146