Logarithmic Differentiation

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Differentiate x to the power x. One student uses the power rule and writes x times x to the power, x minus one. Another student uses the exponential rule and writes x to the power x, times log x. So, who is right? Stay with me, because the answer will surprise you. Today's tool is logarithmic differentiation, and questions on it come up often in JEE papers. So, let's begin.

0:33Why the usual rules fail

So, why is there a doubt at all? Look at the power rule. d by d x of x to the power n equals n times x to the power, n minus one. The d by d x in front means the derivative with respect to x, and the small raised n is the power. But this rule works only when the power n is a constant, a fixed number. Okay? Now the exponential rule, for a to the power x, the one in the question on screen. Tell me, what must stay fixed for that rule to work? The base a, exactly! d by d x of a to the power x equals a to the power x, times log a, only when the base is a constant. These letters, l n, mean the natural log, and we read l n a as log a. Clear? Now look at x to the power x. The base is x, and the power is also x. Both are changing! So the power rule fails, and the exponential rule fails too. We need a new tool. With me?

1:37The trick: take log, differentiate, multiply by y

So here is the trick. Logs are very good at one job. What does a log do to a power? It brings the power down, yes! Call our function y, with x greater than zero, because log x needs a positive x. Take log on both sides: log y equals x times log x. See, the power x came down, and now it is just a product. Right? Now differentiate both sides with respect to x. Careful: y itself is a function of x. So what is the derivative of log y? One over y, times d y by d x, by the chain rule. Correct! And d y by d x is the derivative of y with respect to x. On the right, the product rule: one times log x, plus x times one over x. The dots mean times. Got it? Now the last move, the one students forget: multiply by y. d y by d x equals y times, log x plus one, and y is x to the power x. So the answer is x to the power x, times one plus log x. That's it! And now our puzzle. Open the bracket: x to the power x, plus x to the power x times log x. The first part is the power-rule answer, because x times x to the power, x minus one, is just x to the power x. The second part is the exponential-rule answer. So each student was half right, and the true answer is the sum of both. Surprised?

3:14The general formula for u to the power v

Now the same thing in general. Let y equal u to the power v, where u and v are both functions of x, and u is positive, so that log u makes sense. Okay? Take log: log y equals v times log u. This double arrow means implies. Differentiate: y prime over y equals v prime log u, plus v u prime over u. The small dash is called prime, so y prime is just d y by d x. Right? Multiply by y, which is u to the power v. So, d by d x of u to the power v equals u to the power v, times the bracket, v prime log u, plus v u prime over u. This is the formula in your course. Write it down, okay? And here is a topper's trick. Open that bracket and you get two parts: the power rule with v held fixed, plus the exponential rule with u held fixed. Exactly what we saw for x to the power x. Clear?

4:14Long products and quotients

The same trick has a second use: long products and quotients. The product rule on five factors? Painful, right? So, what does a log do to a product? It turns it into a sum, exactly! And a quotient into a difference. So for y equals f times g, over h, all positive, log y equals log f, plus log g, minus log h. Okay? Now differentiate each log. y prime over y equals f prime over f, plus g prime over g, minus h prime over h. Each piece is really simple! With me? Then y prime equals y times that bracket; the three dots stand for the bracket. Look, this is the biggest trap in this topic. Students find the bracket and stop. But that is y prime over y, not y prime! Never forget this last y. Got it?

5:11Example 1: y'(0) for (x+1)(x+2)(x+3)/((x+4)(x+5))

Alright, example one, from the course. y equals x plus one, times x plus two, times x plus three, all over x plus four, times x plus five. Find y prime of zero, the derivative at x equals zero. Pause the video and try it. Done? Okay, let's check. Near x equals zero every factor is positive, so take log. Each factor has derivative one, so y prime over y equals one over x plus one, plus one over x plus two, plus one over x plus three, minus one over x plus four, minus one over x plus five. Okay? Now put x equals zero. y is one times two times three, over four times five: six over twenty, which is three over ten. Right? And the bracket: one, plus one half, plus one third, minus one fourth, minus one fifth. Take sixty as the common denominator, and you get eighty three over sixty. Clear? So the answer is eighty three over sixty, right? No, no, careful! That is y prime over y. Multiply by y, which is three over ten. Three over ten, times eighty three over sixty, is eighty three over two hundred. That is the answer. See how the trap works?

6:35Example 2: dy/dx when x to the power y = y to the power x

Now a harder one, JEE level. x to the power y equals y to the power x. Find d y by d x. Pause and try. Hint: take logs first, then differentiate implicitly, as in the previous concept. Done? Okay, let's check. Both sides have a variable power, so with x and y positive, take log. Log of x to the power y is y log x, and log of y to the power x is x log y. Right? Now differentiate the left side, y times log x. Careful, y is a function of x. What do you get? Product rule: y prime log x, plus y times one over x. Yes, exactly! And on the right, x log y gives log y, plus x times one over y, times y prime. Don't forget that y prime from the chain rule. Okay? Now collect the y prime terms on one side. y prime times the bracket, log x minus x over y, equals log y minus y over x. With me? Divide, and clear the small fractions by multiplying top and bottom by x y. d y by d x equals y times, x log y minus y, over x times, y log x minus x. That is the course answer. Perfect!

8:02Quick revision: take log first, then multiply by y

So, quick revision. Variable base and variable power: take log first. Long products and quotients: log turns them into sums and differences. And the last step, always: multiply by y. Now go and try the practice problems on this page. All the best!