 Hello. Namaste. Myself, MS Bhasragam, I stand professor, department of community design sciences, Palchandrista of Technology, Solapur. In this session, we are going to discuss successive differentiation part 1. Lowering outcome. At the end of this session, students will be able to find nth derivative of algebraic function. Now, first of all, we will see about successive differentiation. What is successive differentiation? When a function has been differentiated once, the result may be again differentiated and the process repeated indefinitely is called successive differentiation. And notation for the successive derivatives of y equal to f of xr. We are denoting the first derivative by dy by dx, second derivative by d square y by dx square, third derivative by d cube y by dx cube and so on. nth derivative can denote d raised to n y by dx raised to n. Or in short, for the shake-up convenience, we can make use the symbols y sub x1, that is for dy by dx, y sub x2 for d square y by dx square, and y sub x3 for d cube y by dx cube and so on. For nth derivative, we can denote y sub xn. Or also, we can denote f dash of x, f double dash of x, f triple dash of x, and so on, f raised to n of x for writing the derivative of the function y equal to f of x. Now, first of all, we will see the nth derivative of some elementary functions in that. First one is algebraic function, that is y equal to ax plus b raised to m. Now, differentiating y equal to ax plus b raised to m with respect to x, we get y1, that is first derivative of y with respect to x. Here, we are denoting y sub x1, means dy by dx, is equal to derivative of ax plus b raised to m, is m into ax plus b raised to m minus 1 into derivative of ax plus b, that is a. Similarly, differentiating this first derivative with respect to x, we can denote y sub x2, means second derivative, that is m into m minus 1 into ax plus b raised to m minus 2 into a into a, that is what is square. Similarly, differentiating the second derivative with respect to x, we get the third derivative, that is y sub x3, which is equal to m into m minus 1 into m minus 2 into ax plus b raised to m minus 3 into a cube. Now, after generalizing, after n differentiation, we can write yn, that is comparing y1, y2 and y3, we can write the n derivative, which is equal to m into m minus 1 into m minus 2 and so on, and there will be m minus of n minus 1, that is comparing here first, second and third derivative. Here, we can write m minus of n minus 1 into a raised to n into ax plus b raised to m minus n, if n is less than m. Let us call this as equation number one, here we get three cases, this is the derivative, if odd of the derivative is less than m, if m is equal to n, then we get yn is equal to n into n minus 1 into n minus 2 and so on, we get 3 into 2 into 1 into a raised to n and which is you can write n factorial into a raised to n, that is substituting m equal to n in the above result, we get yn equal to n into n minus 1 into n minus 2 and so on, 3 into 2 into 1 into a raised to n, that we can write n factorial into a raised to n. And if n is greater than m, that is if order of the derivative is greater than m, then nth derivative of the function will be ax plus b raised to m will become 0, that is yn is 0, this is the first case. Now second case, if y equal to ax plus b raised to minus m, or we can write y equal to 1 upon ax plus b raised to m. Now changing the sign of m in the above result, that is in equation number one, here just we have to change m by minus m in equation number one, see here equation number one, in this one we have to replace m by minus m, that what we get minus m minus m minus 1 minus m minus 2 and so on, we get here is yn equal to minus m into minus m minus 1 into minus m minus 2 and so on, minus m minus n plus 1, bracket complete into a raised to n into ax plus b raised to minus m minus n. Now here minus sign, taking the minus sign common it is a multiple and we can write this as a minus 1 raised to n into m into m plus 1 into m plus 2 and so on, here you have taken minus 1 common from this one. Therefore, we would get m plus n minus 1 into a raised to n into ax plus b raised to minus m minus n. Now here we will simplify the result for the sake of simplicity, we will adjust the term and we will write the n derivative in simplified form, which is equal to c, we will write minus 1 raised to n reversing the order m plus n minus 1 into m plus n minus 2 and so on, m plus 1 into m just by reversing the order and here m minus 1 into m minus 2 and so on, 3 into 2 into 1, here we have taken this, we have multiplied by extra term and we are dividing by the same term, that is m minus 1 into m minus 2 and so on, 3 into 2 into 1 into a raised to n by ax plus b raised to m plus n. Now we will write this in simplified form, keeping minus 1 raised to n as it is, here we have taken m plus n minus 1 and so on up to 3 into 2 into 1, this complete expression we can write as m plus n minus 1 factorial and denominator m minus 1 into m minus 2 and so on, 3 into 2 into 1, we can write m minus 1 factorial into a raised to n divided by ax plus b raised to m plus n, let us call this as equation number 2, that is here just we have simplified here multiplying and dividing by the same quantity, that is m minus, here we have multiplied m minus 1 into m minus 2 and so on up to 1 and also we have divided and we have written the result in simplified form. Now third case, if y equal to 1 upon ax plus b then yn is equal to minus 1 raised to n into n factorial into a raised to n upon ax plus b raised to n plus 1. Now to get this result, putting m equal to minus 1 equation number 1, that we already seen for equation number 1 general case, in that you have to substitute m equal to minus 1, that is y equal to ax plus b raised to minus 1, which you can write 1 upon ax plus b. Then from equation number 1, you get yn is equal to minus 1 into minus 2 into minus 3 and so on minus n into a raised to n into ax plus b raised to minus 1 minus n and here taking minus 1 common, we will get minus 1 raised to n into 1 into 2 into 3 and so on n into a raised to n into ax plus b raised to minus 1 minus n. Therefore, we can write the n derivative of this as yn is equal to minus 1 raised to n into n factorial that is 1 into 2 into 3 into n and so on n, we can write n factorial and a raised to n taking this term in the denominator, we can write ax plus b raised to n plus 1, this is what equation number 3. Now, these are the three different formula of finding the n derivative of algebraic function. Now, pause the video for a while and write down the n derivative of 1 upon 3x plus phi o. I hope you have completed. Let y equal to 1 upon 3x plus phi o. Therefore, yn is equal to minus 1 raised to n into n factorial into 3 raised to n divided by 3x plus phi o raised to n plus 1. Just now, you have seen if y equal to 1 upon ax plus b then yn equal to minus 1 raised to n into n factorial into a raised to n divided by ax plus b raised to n plus 1. Here, a is if you compare this 1 upon 3x plus phi o, here a is 3 and b is phi o. This is what the n derivative of the given function 1 upon 3x plus phi o. Now, you will see the example. Find the nth derivative of x plus 3 upon x minus 1 into x plus 2. See, this is an algebraic expression. Let y equal to x plus 3 upon x minus 1 into x plus 2. Here, we cannot write the n derivative of this function directly. Therefore, we have to make use of partial function to express this function in the form of 1 upon ax plus b. Let x plus 3 upon x minus 1 into x plus 2 is equal to a upon x minus 1 plus b upon x plus 2 because you are already familiar with how to take the partial fraction for a given function. Here, the degree of the numerator is 1 and degree of the denominator is 2. Therefore, and two factors are linear and distinct. Therefore, we can write x plus 3 upon x minus 1 into x plus 2 equal to a upon x minus 1 plus b upon x plus 2. Now, we have to multiply both side by x minus 1 into x plus 2. We get therefore, x plus 3 equal to a into x plus 2 x minus 1 will get cancelled plus b into x minus 1 and x plus 2 will get cancelled. Now, here our aim is to find a and b. Therefore, we have to choose the suitable values of x. Now, to find a, putting x equal to 1 both side or left hand side 1 plus 3 that is 4 and putting x equal to 1, what you get a into 1 plus 2 that is 3, 3a and this second term will become 0. Therefore, what you get a equal to 4 equal to 3a therefore, a equal to 4 by 3. Now similarly, putting x equal to minus 2 to find b both side putting x equal to minus 2 on left hand side you get minus 2 plus 3 that is 1 and putting x equal to minus 2 first term will become 0 b into minus 2 minus 1 that is minus 3 b therefore, b equal to minus 1 by 3. Now, substituting the values of a and b here we get the given function the given function can be written as y equal to a that is 4 by 3 into 1 upon x minus 1 and b is minus 1 by 3 that is minus 1 by 3 into 1 upon x plus 2 and now here we have expressed the given function in the form of 1 upon ax plus b and now we are able to write the n derivative of the given function. See therefore, y n equal to now we can compare the given function with 1 upon ax plus b if y equal to 1 upon ax plus b then y n equal to what we have minus 1 raise to n into n factorial into a raise to n upon ax plus b raise to n plus 1. Now here you comparing with 1 upon ax plus b here a is 1 and b is minus 1 in case of first one. See here you can compare here in this first case a is 1 b is minus 1 and in second term a is 1 and b is 2. Therefore, n derivative is 4 keeping 4 by 3 as it is and the n derivative of is minus 1 raise to n into n factorial divided by x minus 1 raise to n plus 1 applying this formula. Similarly, minus 1 by 3 as it is and n derivative of 1 upon x plus 2 is minus 1 raise to n into n factorial divided by x plus 2 raise to n plus 1 because here a raise to n term is 1 in both the cases the value of a is 1 now which is equal to now we are taking 1 by 3 common minus 1 raise to n common and n factorial common that is 1 by 3 into minus 1 raise to n into n factorial into bracket we get 4 upon x minus 1 raise to n plus 1 minus here you get 1 upon x plus 2 raise to n plus 1 references higher engineering mathematics by Dr. B. S. Gravel. Thank you.