Leibniz's Rule

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle to start. Take the integral from zero to x squared of sine t, d t, and find its derivative. Most students integrate first, put in the limits, and only then differentiate. Now, what if the inside were sine of t squared instead? That cannot be integrated in the usual way. So is the derivative impossible? Not at all, because with Leibniz's rule it takes one line, and you never integrate. Questions like this come up often in JEE, so let's begin.

0:37Upper limit x: the fundamental theorem

So, start with the simplest case. Capital F of x is the integral from a to x of f of t, d t. The long S is the integral sign, the lower limit a is a fixed number, and the upper limit is x. The t inside is only a dummy variable, okay? Now tell me, what is F prime of x? The little dash, the prime, means the derivative with respect to x. Just f of x, yes, exactly! If f is continuous, then F prime of x equals f of x. This is the fundamental theorem of calculus; NCERT calls it the first fundamental theorem. So simply put t equals x in f of t, but do remember the condition: f continuous. Right?

1:23Leibniz's rule: limits that depend on x

Now the real thing. What if both limits move with x? Here the lower limit is g of x, and the upper limit is h of x. The d over d x in front means: differentiate the whole integral with respect to x, okay? From the last rule, the upper limit gives f of h of x. But the upper limit is now h of x, not plain x. So what extra factor must come in? The derivative of h, yes! That is just the chain rule. So the answer is f of h of x, times h prime of x, minus f of g of x, times g prime of x. The lower limit always comes with a minus sign. With me? Here is the easy way to remember it. Upper limit: put it into f, then multiply by its derivative. Okay? Lower limit: do exactly the same, and subtract. Oh no, careful here! The common mistake is to forget h prime, or to drop the lower limit term. Got it? And the conditions, as always: f continuous, and the limits g and h differentiable functions of x. Right?

2:32When x is also inside the integrand

Now a trap that catches many students. Look at g of x, the integral from zero to x of x minus t, times f of t, d t. See? The x also sits inside the integrand, not only in the limit, okay? So, just put t equals x inside. That gives x minus x, times f of x, which is zero. So g prime of x is zero, right? No, no, careful here! That rule works only when x sits in the limits alone. So first take x outside: x times the integral of f of t, minus the integral of t times f of t. Inside the integrals, x is now a constant. Clear? Now differentiate. The first piece is a product, so the product rule gives the integral of f of t, plus x times f of x. And the second piece, with its minus sign, gives minus x times f of x, by our rule. With me? The plus x f of x and the minus x f of x cancel. So g prime of x is the integral from zero to x of f of t, d t. Differentiate once more, and g double prime of x, with two dashes, is f of x. Got it?

3:43Limits of the form 0/0: L'Hôpital meets Leibniz

Last idea. Look at this limit: as x tends to zero, one over x cube, times the integral from zero to x of sine of t squared, d t. The word lim means limit, and the arrow means tends to, okay? And this is the integrand from our hook, the one we cannot integrate. So what happens to the top and the bottom as x tends to zero? Both go to zero, correct! The integral from zero to zero is zero, and x cube is zero too. So it is the zero over zero form. Right? So we use L'Hôpital's rule: differentiate the top and the bottom separately. The top, by our rule, gives sine of x squared. The bottom gives three x squared. Okay? Now take one third outside. Sine of x squared, over x squared, tends to one. So the limit is one third, and we never integrated at all! Got it?

4:39Example 1: d/dx of integrals of sin t and ln t up to x2

Now Example one. The first part is our puzzle. Find the derivative of the integral from zero to x squared of sine t, d t. Second, find the derivative of the integral from x to x squared of l n t, d t, for x greater than one. Here l n means the natural log. Pause the video and try it first. Done? Okay, let's check. In the first part, the lower limit is zero, a constant, so it gives nothing. The upper limit is x squared. Put it into sine, and multiply by its derivative, two x. The dot means times. So the answer is two x sine x squared. Puzzle solved, in one line! Now the second part. The upper limit x squared gives natural log of x squared, times two x. Okay? But this time the lower limit is x, not a constant. So what does it give? Minus natural log of x, times one, yes! Put x into log t, multiply by the derivative of x, which is one, and attach the minus sign. Right? Now simplify. Since x is positive, natural log of x squared is two log x, so the first term becomes four x log x. In total, four x log x minus log x. Clear? Take log x common, and we get four x minus one, times natural log of x. That's it! If you forget the lower limit, you lose the minus log x term.

6:07Example 2: limit of integral of cos t2 over x sin x is 1

Now Example two, a little harder. Evaluate the limit, as x tends to zero, of the integral from zero to x squared of cos of t squared, d t, over x sine x. Pause the video and try it first. Done? Okay, let's check. Before we use L'Hôpital's rule, what must we check first? The form, exactly! At x equals zero the integral runs from zero to zero, so the top is zero, and x sine x is zero too. So it is zero over zero, and L'Hôpital is allowed. Okay? Now the top. The upper limit is x squared. Put it into cos of t squared, and multiply by its derivative. What do you get? Two x, times cos of x to the power four. Very good! Careful here: t squared becomes x squared, squared, which is x to the power four. And the bottom, x sine x, by the product rule gives sine x plus x cos x. Right? Now divide the top and the bottom by x. The top becomes two cos of x to the power four, and the bottom becomes sine x over x, plus cos x. Got it? As x tends to zero, cos of x to the power four tends to one, sine x over x tends to one, and cos x tends to one. So we get two times one, over one plus one, which is one. Perfect!

7:32Quick revision: Leibniz rule, x inside the integrand, 0/0

So, quick revision. With upper limit x, the derivative of the integral is just f of x, for f continuous. Leibniz: upper limit in, times its derivative, minus the same for the lower limit. If x sits inside the integrand, take it outside first. And an integral over a power of x giving zero over zero? L'Hôpital with Leibniz. Now go and try the practice problems on this page. All the best!