Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Differentiate sine inverse of, two x times the square root of one minus x squared, for x between one over root two and one. Many students put x equals sine theta, and write in ten seconds: two over the square root of one minus x squared. Looks perfect, right? But it is wrong! The correct answer has a minus sign. So where does that minus sign come from? This is Set two of our mixed problems on limits, continuity and differentiability, and traps like this come up often in JEE, so let's begin.

0:39Set 2: what is inside

So, here is what Set two holds. Eight solved problems in the JEE Main pattern, all mixed up, just like the real exam. Okay? They cover limits, continuity, derivatives, rates of change and increasing functions. Right? Today, two tools, then two problems from the set: an easy one first, and then our puzzle. With me?

1:00Tool 1: differentiable at a point

First tool, from Chapter four. f prime of a: the small dash is read prime, and it means the derivative at a. It is the limit, as h tends to zero, of f of a plus h, minus f of a, all over h. The arrow means tends to, and lim means limit. This limit must exist and be finite. Okay? Now think. This little h can come from the right, positive, or from the left, negative. So what must happen for this limit to exist? The two sides must agree. Yes, exactly! L f prime of a is the left hand derivative, and R f prime of a the right hand derivative. f is differentiable at a if and only if they are equal and both finite; the double arrow means if and only if. Clear? Look at the left picture: y equals the absolute value of x, with two vertical bars. It is continuous at zero, no break. But is it differentiable at zero? No! R f prime of zero, the slope from the right, is one; L f prime of zero, the slope from the left, is minus one. They disagree: a sharp corner, and no derivative at zero. Remember this corner for Example one. Got it? So, differentiable implies continuous; the single arrow means implies. But not the reverse: the absolute value of x is continuous at zero, yet not differentiable. The right picture, the cube root of x, with a small three on the root sign, is continuous too, but has a vertical tangent at zero, an infinite slope. Not differentiable either. With me?

2:43Tool 2: inverse sine after a substitution

Second tool, from Chapter five. Sine inverse of x: the small minus one up there is not a power, it means inverse. Its derivative, d by d x, is one over the square root of one minus x squared. Okay? When a question has the square root of one minus x squared, the course says: substitute x equals sine theta. Theta is the Greek letter that looks like an O with a line across it. Right? Now a question. Take sine inverse of sine phi; phi is another Greek letter, for an angle. Inverse undoes the function, so this is always phi, right? No, not always! Sine inverse of x is the angle between minus pi over two and pi over two whose sine is x. The square brackets mean both ends are included. So sine inverse never gives an angle outside this interval. Clear? So, sine inverse of sine phi equals phi only when phi belongs to that interval; the curved symbol like an E means belongs to. Outside it, sine inverse gives a different angle with the same sine. Got it? So always check the interval before you cancel sine inverse against sine. Skip it, and you get our puzzle's wrong answer. Okay?

4:00Example 1: is x|x| differentiable everywhere? f'(x) = 2|x|

Now, Example one from the set, an easy one. Is f of x equals x times the absolute value of x differentiable everywhere? Find f prime of x. Pause the video, and try it first. Done? Okay, let's check. Many students say: the absolute value of x has a corner at zero, so x times the absolute value of x must also fail there, right? No, no, wait! Check it. For x greater than zero, the absolute value of x is just x. So f of x is x squared, and f prime of x is two x. Okay? For x less than zero, the absolute value of x is minus x. So f of x is minus x squared, and f prime of x is minus two x. Right? Now at zero, use the definition. f of zero is zero, so f of h, minus f of zero, over h, is h times the absolute value of h, over h. The h cancels, leaving the absolute value of h, which tends to zero from the right and from the left. Clear? So L f prime of zero equals R f prime of zero, and both are zero. So f prime of zero is zero: differentiable at zero too! The extra factor x flattened the corner. With me? Put it together: two x for positive x, minus two x for negative x, and zero at zero. All three are two times the absolute value of x. So yes, differentiable everywhere, and f prime of x is two times the absolute value of x. Perfect!

5:41Example 2: sin inverse 2x root(1 - x2), x in (1/root 2, 1)

Now, back to our puzzle: Example two, the harder one. Differentiate sine inverse of, two x times the square root of one minus x squared, for x between one over root two and one. Pause the video and try. Hint: check the interval. Done? Okay, let's check. Put x equals sine theta. Sine of pi over four is one over root two, and sine of pi over two is one. So theta lies between pi over four and pi over two; round brackets mean the ends are not included. Okay? Inside, the square root of one minus sine squared theta is cos theta, positive here since theta is less than pi over two. So the inside is two sine theta cos theta, which is sine two theta. Right? So the function is sine inverse of sine two theta. Sine inverse and sine cancel, so the function is two theta, and the derivative is two over root of one minus x squared, right? Oh no, careful! That is the trap. Theta is between pi over four and pi over two, so two theta is between pi over two and pi. That is outside the interval from minus pi over two to pi over two, so the function is not two theta. Clear? So we need an angle inside the interval with the same sine as two theta. Which one? Pi minus two theta. Yes, exactly! Sine of pi minus two theta equals sine two theta, and pi minus two theta lies between zero and pi over two, inside the interval. Got it? So the function is pi minus two theta, that is, pi minus two sine inverse x. With me? Now differentiate. Pi is a constant, so its derivative is zero. The derivative is minus two over the square root of one minus x squared. There is our minus sign! Puzzle solved. The quick answer had the wrong sign, and only the interval check catches it. Perfect!

7:37Quick revision: left and right derivatives, interval check

So, quick revision. One: f is differentiable at a only when the left and right derivatives are equal and finite, and then it is continuous there too. Two: x times the absolute value of x is differentiable everywhere, and its derivative is two times the absolute value of x. Three: sine inverse of sine phi equals phi only when phi lies between minus pi over two and pi over two, so check the interval. Now go and try the practice problems on this page. All the best!