Inverse Trigonometric Functions and Substitution

Transcript

0:00Introduction

Hello everyone! Okay, a small challenge before we start. Differentiate tan inverse of two x over one minus x squared. What would you reach for? The quotient rule inside the chain rule? Half a page of algebra, right? Today we do it in one line, with one clever substitution. And a twist: for a very similar function, the derivative is plus two over one plus x squared for some x, and minus that for others. Same function, opposite signs? By the end, you will know why. This comes up often in JEE, so let's begin.

0:37Six derivatives, only three to remember

So, first the formulas. See the small minus one written up high on sin? It is not a power, okay? It means inverse: sin inverse x. And d by d x means the derivative with respect to x. So, d by d x of sin inverse x is one over the square root of one minus x squared, for modulus of x less than one; the two vertical bars mean modulus. Right? Next, tan inverse x. Its derivative is one over one plus x squared, with no square root, so it works for every real x. Simple! Third, sec inverse x. Its derivative is one over modulus of x, times the square root of x squared minus one, for modulus of x greater than one. Students often drop that modulus. Don't! Clear? Now, what about the other three: cos inverse, cot inverse and cosec inverse? Can you guess the derivative of cos inverse x? Yes, exactly! The same as sin inverse, with a minus sign: minus one over the square root of one minus x squared. In the same way, cot inverse copies tan inverse, and cosec inverse copies sec inverse, each with a minus sign. Okay? Why? Here is a topper's way to see it. Sin inverse x plus cos inverse x equals pi over two, a constant. Differentiate both sides, and the two derivatives must add up to zero. So one is just minus the other! Six formulas, only three to learn. With me?

2:17Substitution: spot the shape

Alright, now the real power of this chapter: substitution. When an inverse function has an ugly expression inside, put x equal to a trig function, and the ugliness disappears. First shape: the square root of one minus x squared. Which substitution removes that root? x equals sin theta. Correct! Theta is just the Greek letter for an angle. One minus sin squared theta is cos squared theta, a perfect square, so the root goes away. Okay? Second shape: one plus x squared. Put x equals tan theta, because one plus tan squared theta is sec squared theta. Right? Now think. With x equals tan theta, what does two x over one plus x squared become? Sin two theta! Very good! The double angle formulas in disguise. Two x over one minus x squared is tan two theta, two x over one plus x squared is sin two theta, and one minus x squared over one plus x squared is cos two theta. Three faces of one substitution. Got it? One more shape, where a and b are constants: a cos x minus b sin x, over b cos x plus a sin x. Divide the top and the bottom by b cos x, and let tan alpha equal a over b. Alpha is another Greek letter, a fixed angle. Then the whole thing is tan of alpha minus x, by the tan subtraction formula. Clear?

3:53The trap: check the interval first

Now, the most important warning of this video. After a substitution, we get things like sin inverse of sin phi; phi is one more Greek letter for an angle. Sin inverse and sin undo each other, so the answer is always phi, right? No, no, wait! Let's test it with phi equal to pi. Sin pi is zero, and sin inverse of zero is zero. Zero, not pi! The equals sign with a slash means not equal to. Okay? The reason: sin inverse only gives answers from minus pi over two to pi over two. So sin inverse of sin phi equals phi only for phi from minus pi over two to pi over two, both ends included. That is what less than or equal to means. Right? For tan inverse, the same range, but with the ends left out: phi strictly between minus pi over two and pi over two. With me? And for cos inverse, phi must lie from zero to pi, both ends included. Clear? And when phi falls outside? Use an identity to bring the angle back inside; for an angle a little above pi over two, sin of pi minus phi equals sin phi does the job. So, always check the interval before you cancel. We need this in example two. Okay?

5:13Example 1: tan inverse 2x/(1 - x2) using x = tan theta

Now, remember the challenge from the start? Let's finish it. Differentiate tan inverse of two x over one minus x squared, for modulus of x less than one. Pause the video, and try it first. Hint: which face of tan theta is this? Done? Okay, let's check. Two x over one minus x squared is the tan two theta face, right? So put x equals tan theta, and we get tan inverse of tan two theta. Now, can we cancel straight away, and write two theta? Not before checking the interval! Modulus of x less than one means theta lies between minus pi over four and pi over four; the double arrow means implies. So two theta lies between minus pi over two and pi over two. Okay? That is the safe range of tan inverse, so now we cancel. The function is two theta, and theta is tan inverse x. So, two tan inverse x. Right? And its derivative is two over one plus x squared. That's it! No quotient rule, no half page of algebra. One line. Perfect!

6:26Example 2: sin inverse 2x/(1 + x2), sign flip for x above 1

Now a harder one, also from the course. Differentiate sin inverse of two x over one plus x squared, first for modulus of x less than one, then for x greater than one. Pause and try. Hint: part b hides the trap. Done? Okay, let's check. Two x over one plus x squared is the sin two theta face. So put x equals tan theta: sin inverse of sin two theta. Got it? Part a, modulus of x less than one. Just like example one, two theta lies between minus pi over two and pi over two. Safe range, so we cancel: the function is two tan inverse x, and the derivative is two over one plus x squared. Easy! Now part b, x greater than one. Same expression, so the same answer, right? Oh no, careful! Check the interval. x greater than one means theta lies between pi over four and pi over two, so two theta lies between pi over two and pi. Outside the range of sin inverse! We cannot cancel. With me? So we use our rescue line. Sin two theta equals sin of pi minus two theta, and pi minus two theta lies between zero and pi over two, safely inside. Clear? So sin inverse gives back pi minus two theta, which is pi minus two tan inverse x. Pi is a constant, so the derivative is minus two over one plus x squared. There it is, the twist from the start! The sign flips when x crosses one, all because of the interval. Very good!

8:10Quick revision: 3 derivatives, 2 substitutions, range check

So, quick revision. Learn three derivatives; cos inverse, cot inverse and cosec inverse just add a minus sign. For the square root of one minus x squared, put x equals sin theta; for one plus x squared and the two theta faces, put x equals tan theta. And before you cancel sin inverse of sin phi, check that phi is in the range. Now go and try the practice problems on this page. You are ready. All the best!