Practical lesson 2. Higher-order derivatives and differentials
At this lesson, we will learn how to find higher-order derivatives and differentials. To begin with, we will try to find the first-order differential, then we will consider higher orders. In general, the problem of calculating the second- and third-order derivative and higher is not difficult at all.
I will not go into detail about how it is calculated, I think, you can easily handle it. For example, there is a function given, your task is to find a third-order derivative. We follow the steps sequentially. We find the first-order derivative, it is more convenient to do this by converting f to such a power form. Having obtained the first-order derivative, we find the second-order derivative. Then, having received the answer, we differentiate again, and find the third-order derivative. The procedure is not complicated, and if you know how to find a first-order derivative, you will not have any problems with the rest either.
Let’s speak about the differential. You can find it using two ways. What is meant here?
The first method is the simplest and most logical. Let’s write it down using the formula: df(2)=f`(2)*dx. Everything is quite simple, remember, on the left, there is a differential, and on the right, there is a derivative. The derivative and differential differ by dx multipliers only. This is a very important note when you write a differential – there is d on the left, and there should also be a differential sign on the right. The derivative at point 2, the derivative y` = 2x-2, and the derivative at point 2 will be equal to 2. Therefore, the differential at point 2 is 2dx. (df (2)=2dx).
Now let’s try to find this differential using the second way, using the definition: the differential is the main linear part of the increment of the function. So you need to act incrementally. Increment formula: we give the increment argument ∆x and see how the value of the function changes: the value at point 0 (2+∆x) and subtracts the value at the original point 2. Calculate the values of the function using the formula: ((2+∆х)2)– (2(2+∆x) +1. The value at point 2, let’s count: 4 – 4+1 = 1. 1s can be removed immediately.
Opening the brackets, we get: 4+4∆x+ ∆x^2-4-2 ∆x. Let’s change all terms: we have eliminated 4s and all terms with ∆x. (2∆x+∆x^2) Look, the increment of the function turned out to be represented as the sum of two terms, 2∆x is the main part, and ∆x^2 is an infinitesimal part of a higher order than ∆x. The main linear part is the differential. Let’s compare, the answer is 2dx.
So, the differential of the function and we remember ∆x=dx. Therefore, the differential of the function f at point 2 is 2dx - .This is the main linear part of the function increment.
Sometimes there is a problem that can be solved, but you do not want to solve it at all, and not because it is difficult, but because it is long. Find the twentieth-order derivative of the function. In this case, we must consistently perform twenty steps, to differentiate may not be difficult, but long. What is the smart thing to do in this case? Try to derive the nth derivative, the formula for it, and then specify it. And so, y` = 1/x. For convenience, let’s write x^(-1) to make it easier for us to find the following derivatives. Y`` = -x^(-2), y```=2x^(-3). Let’s find another one and try to analyze it. YIV = 2*(-3)*x^(-4), we will not move to the twentieth derivative.
Let’s try this step. Let’s try to find what these formulas have in common: the nth derivative, it is immediately clear that x is in the power (-n). What are cofactors? If the degree is even and the order of the derivative is the degree of X, it will be the minus sign, “+” will be for odd ones. So we write: (-1) ^(n+1). So, the sign is clear. What will happen with pure cofactors? We took the sign into account. In the next step, we add 4, and the product of natural numbers is the factorial. The fourth power, and products up to three, which means (n-1).
Yn= ((-1)^(n+1))* (n-1)!* x^(-n).
We have an assumption. I can’t say that what we got is true. This is our hypothesis. And we prove the formula by mathematical induction. In the first step, for n=1, we check that the base is correct, that we did not make a mistake. Yn=(x^(-1))(0!)(-1)2 - plus sign. The base is executed for n=1 – this is true. Let’s make a transition from n → n+1, we find an inductive transition, we find the n+1 derivative, as the derivative of the nth. And let’s see if the specified formula is fulfilled. We find the derivative: y(n+1) = ((-1)^n+2)* n!* (x^(-n-1)). To make it clearer that the formula is confirmed, y(n+1) = ((-1)^n+1+1)*n!* (x^-(n+1)). We check, if we substituted n+1 instead of n, would we get this formula? Yes! We actually get this formula. So our assumption was correct. The method of mathematical induction allows us to conclude that the formula is true for any n-natural. So, we find the twentieth derivative using the formula. Y20= ((-1)^21)*19!*(x^-20). We write down: y20=-19!/(x^20). We have solved the problem!
The problem can be initially written as follows: find the nth derivative. The situation here is quite simple. Let’s take a look and repeat the reasoning again: y`=3e^(3x+1); y``= (3^2) *e^(3x+1). And we already guess that at each step, a multiplier of 3 is simply added. Therefore, the hypothesis arises: the nth derivative is (3^n) *e^(3x+1). If n=1 – the formula is correct, the base is executed. The inductive transition. We find the (n+1)th derivative by differentiating the nth: yn+1=(3^n*n^(3x+1))`=3n+1*e3x+1. So, the answer is obtained by the formula: yn= (3^n)*e^(3x+1).
Let’s remember what difficult situations arise? A power-exponential function is given. And the task is to find derivatives and differentials of the first and second order. Let’s recall the formula for logarithmic differentiation: y`= y*(lny)`. I’ll write it down right here. This is y` = (x^ln(x))*(ln(x)*ln(x))`= (x^ln(x))*(ln2(x))`= xlnx*2lnx*1/x. That is the answer. Of course, two can be written before all cofactors. You can immediately write the differential: dy=y`dx, and all combined: dy= 2xlnx-1*lnx*dx.
And now let’s solve the problem of the second order. I specifically saved this entry to make it easier for us to find the second-order derivative. Even if in this form: here we have 3 cofactors, so the derivative of this function: we rewrite the generalization of the formula (y``=xlnx*2lnx*1/2). The derivative of the product of two functions when we generalize it to the product of any finite number of cofactors. What are we doing? We rewrite it since there are 3 cofactors, we rewrite three times. ((xlnx)` *2lnx*1/2))+(xlnx*(2lnx)`*1/2)+ (xlnx*2lnx*(1/2)`). We put the primes above the first, second, and third ones.
And then there will be practically a record of the answer, perhaps cumbersome, and we will not finish it. Look: x^lnx is a given function, so instead of the first multiplier, we can write the equation: 2xlnx-1*lnx, it is the derivative of the first one. And we have: 2lnx, so 22xlnx-2*ln2x –the derivative of the logarithm (1/x) here. 22xlnx-2*ln2x+2xlnx*(1/x2 ). Here, the derivative will be -1/x^2, so we write -(1/x). yII = 22xlnx-2*ln2x+2xlnx*(1/x2 ) - 2xlnx*lnx*(1/x^2). You can still convert the answer, but in fact, you can leave it in this form. I will not write down the formula in the full form of a second-order differential. We’ll just remember the formula. The second order differential is found as the second derivative multiplied by dx^2. So what do we need to do? We enclose all long expressions in parentheses and multiply by dx^2. And the problem is solved. A bit cumbersome, but everything is solved and everything is available.
Another generalization. What generalization do we have for the formula?
The first generalization is when we consider, and we have just done it. The first generalization is based on the number of cofactors. And in the previous problem, this generalization was considered. In this problem, the generalization of the formula concerns the order of the derivative. So, the derivative of the product of two functions, but not of the first order, but of the third one. The formula for the third order in this case looks like this: (uv)```=u```v+3u``v`+3u`v``+uv```. To use the formula, what do we need to know? For each of the functions x and lnx, you need to know all the derivatives from the first to the third order, inclusive. So I write: u=x, and I find all derivatives up to and including the third order. Similarly, I write: v=lnx, and I find all derivatives up to the third order inclusive. u`=1; u``=0; u```=0. V`=1/x; v``=-1/x^2; v```=2/x^3.
Let’s see. So, the answer is obtained immediately. (xlnx)```. The derivative u```= 0 and the gerivative u``= 0, thus, we have 2 cofactors left. (xlnx)```=3*1-(1/x^2) +x*2/x^3=(-3/x^2) +2/x^2=-1/x^2. We could have done this as follows: we find the first-order derivative, so differentiating the answer – the second one. And the answer would be the same. Our task may not be the most optimal solution, but it is simply to show how the generalization formula works and how it can be used. Perhaps, direct differentiation led to an answer even faster, but in some cases, especially when there are no zero derivatives, as here, this solution seems to be optimal.
And let’s see how we write coefficients. In fact, these are coefficients that you know from abbreviated multiplication formulas since high school. For the second derivative, these are the coefficients from the formula (a+b) ^2; for the third derivative, (a+b) ^ 3; for the fourth – (a+b) ^4, and so on. You can use formulas for the number of combinations, there are other techniques, but I think this is enough for us for now.