Higher Derivatives

Transcript

0:00Introduction

Hello everyone! Okay, a little game first. Differentiate sine x: you get cos x. Again: minus sine x. Again: minus cos x. And once more: sine x, back where we started! By the end of this video, you will write any derivative of sine x in one line. We will also catch a parametric trap that fools many students. Higher derivatives come up often in JEE papers, so let's begin.

0:31The second derivative: differentiate once more

So, a higher derivative is nothing new. Differentiate d y by d x once more, with respect to x, and you get the second derivative. We write it as y double prime, with two small dashes, or as d squared y by d x squared. See the small twos? One sits on the d on top, one after the d x at the bottom. Okay? And in function notation, it is f double prime of x: the derivative of f prime of x. Right? Now a quick question. Is d squared y by d x squared the same as d y by d x, the whole squared? No, not at all! d squared y means differentiate twice. The square means multiply the first derivative by itself. This crossed equals sign means not equal to. Clear?

1:21Parametric curves: the second derivative

Now, parametric curves: x and y are both functions of a third variable t, the parameter. Recall: d y by d x equals d y by d t, divided by d x by d t. The slanted line means divided by. Okay? So, for the second derivative, many students take y double prime of t, the second derivative of y with respect to t, and divide by x double prime of t. Looks natural, right? So, does it work? No, no, careful! d squared y by d x squared means differentiate d y by d x with respect to x, not t. So, by the chain rule: d by d t of d y by d x, all over d x by d t. With me? So remember this line: dividing the two second derivatives is wrong. And since we divide by d x by d t, it must not be zero. Clear?

2:17Second derivative of x with respect to y

Next, the same idea the other way round. d x by d y is one over d y by d x, when d y by d x is not zero. Okay? So, is d squared x by d y squared just one over d squared y by d x squared? No! Same mistake as before. We are differentiating with respect to y, so the chain rule comes in. The result: d squared x by d y squared equals minus d squared y by d x squared, over d y by d x, the whole cubed. Look, a minus sign and a cube at the bottom. Write this one down. Got it?

2:57nth derivatives: spot the pattern

Now, nth derivatives. Three patterns save a lot of time. Start with e to the power a x. Each time you differentiate, the chain rule brings down one more a. So, after n times? You get a to the power n, times e to the power a x. Yes, exactly! The small n in round brackets on top means the nth derivative, not the nth power. Simple, right? Now, our sine puzzle. The derivative of sine x is cos x, which is sine of x plus pi over two. Pi is the Greek letter on screen. So each derivative adds pi over two to the angle, and after n derivatives: sine of x plus n pi over two. Put n equals four: two pi, a full turn, so back to sine x. Puzzle solved! Third pattern: one over x plus a. Each arrow means differentiate once more. Minus one over x plus a squared, then two over x plus a cubed, then minus six over x plus a to the power four. So, what pattern do you see? The signs alternate, and one, two, six are one factorial, two factorial, three factorial. Correct! The exclamation mark is read factorial. So the nth derivative is minus one to the power n, times n factorial, over x plus a to the power n plus one, for x not equal to minus a. Okay? And a topper's trick. For one over x squared minus one, first split into partial fractions: one half, times one over x minus one, minus one over x plus one. Now each piece follows the pattern. With me?

4:50Derivative of one function with respect to another

One last idea. The derivative of f of x with respect to g of x is f prime of x over g prime of x, with g prime of x not zero. It is the parametric idea again, with x playing the role of t. Okay? So, quickly: what is the derivative of x squared with respect to x cubed? Two x over three x squared, which simplifies to two over three x. Very good!

5:17Example 1: second derivative of x cubed ln x

Alright, example one. Find y double prime for y equals x cubed, times l n x. The letters l n mean the natural log, and we read l n x as log x. Pause the video and try it. Done? Okay, let's check. First derivative, by the product rule: three x squared log x, plus x cubed times one over x. The dot means times. So, y prime equals three x squared log x, plus x squared. Clear? Now differentiate again. Three x squared log x gives six x log x, and x squared gives two x. So y double prime is six x log x plus two x, right? No, no, careful! Three x squared log x is again a product. The product rule gives six x log x, plus three x squared times one over x. Many students forget this second piece. Then add the two x. Okay? Three x squared over x is three x, and three x plus two x is five x. So, y double prime equals six x log x, plus five x. That's it!

6:38Example 2: d2y/dx2 for x = a cos cubed t, y = a sin cubed t

Now a harder one, JEE level. x equals a cos cubed t, and y equals a sine cubed t. The small three means cos t, the whole cubed, and the same for sine. Find d squared y by d x squared. Pause and try. Hint: use the parametric formula. Done? Okay, let's check. By the chain rule, d x by d t equals minus three a cos squared t, times sine t. And d y by d t equals three a sine squared t, cos t. Right? Now divide. The three a cancels, one sine t cancels, and one cos t cancels. Sine t over minus cos t: so d y by d x equals minus tan t. Nice and clean. With me? Next, differentiate d y by d x with respect to t. The derivative of minus tan t is minus sec squared t. Okay? So the answer is minus sec squared t, right? No, no, careful! We differentiated with respect to t, not x. Divide by d x by d t, which is minus three a cos squared t, times sine t. The minus signs cancel, and sec squared t is one over cos squared t. So, d squared y by d x squared equals one over, three a cos to the power four t sine t, with all of that in the denominator. Perfect!

8:08Quick revision: second, parametric and nth derivatives

So, quick revision. Second derivative: differentiate d y by d x once more. Parametric: differentiate d y by d x with respect to t, then divide by d x by d t. Never divide the two second derivatives. And for nth derivatives, learn the patterns for e to the power a x, sine x, and one over x plus a. Now go and try the practice problems on this page. All the best!