 Hello, Namaste, Myself, MS Bhashrakam, Assistant Professor, Department of Humanities and Sciences, Walchand District Technology, Solapur. In this session, we are going to discuss successive differentiation part two, learning outcome. At the end of this session, students will be able to find nth derivative of algebraic function. See, in the first video, already we have seen how to find the nth derivative of a given function, that is algebraic function, and we have developed the formula, and we have seen the examples. We will see few more examples on finding the nth derivative of algebraic function. Now, first example is, find the nth derivative of 1 upon 6x square minus 5x plus 1. y equal to 1 upon 6x square minus 5x plus 1. To apply the general formula of nth derivative, here first of all, we have to factor, because this is in the denominator of quadratic equation is there. First of all, you have to factorize this as a product of two linear factors. See here, we have 6x square minus 5x plus 1. We can write 6x square minus 2x minus 3x plus 1. Here we have just the middle term, such that here we are registering the middle term, such that their product is plus 6 and sum is minus 5. That is product of minus 2 and minus 3 is plus 6, that is satisfying, and sum of minus 2 and minus 3, that is minus 5. And now, in the first two terms, we are taking here 2x is common, and here 2x into bracket 3x will remain minus 1, and here we are taking the minus sign common, 3x minus 1. Therefore, taking 2x minus 1 common, you get factors are 2x minus 1 into 3x minus 1. Therefore, we can write the given function y equal to 1 upon 6x square minus 5x plus 1 as 1 upon 2x minus 1 into 3x minus 1. Now again, here we will go for partial fraction, because we have to apply the formula of 1 upon ax plus b, let 1 upon 2x minus 1 into bracket 3x minus 1 equal to a upon 2x minus 1 plus b upon 3x minus 1, because both the factors are linear and distinct. By using the property of partial fraction, we have written this expression. Now multiplying both side by 2x minus 1 into 3x minus 1, you get left-hand side will remain 1 is equal to a into 3x minus 1 plus b into 2x minus 1. Now, we have to find the values of a and b, for that we have to choose the suitable values of x. Now to find a, putting x equal to 1 by 2, now we have to put x equal to 1 by 2 here both side, putting we have to put x equal to 1 by 2, left-hand side will be 1 only. Then what you get? 1 is equal to a into bracket 3 by 2 minus 1, that is 1 is equal to 3 by 2 minus 1, that is 1 by 2 a, therefore a equal to what you get 2. And to find b, putting x equal to 1 by 3, you get 1, left-hand side is 1, is equal to b into bracket 2 by 3 minus 1, that is 1 equal to minus 1 by 3 b, therefore the value of b is minus 3. Now, substituting the values of a and b, substituting the values of a and b here, therefore we can write the given expression as y equal to 2 upon 2x minus 1 minus 3 upon 3x minus 1. Now we already seen the n derivative formula, if y equal to 1 upon ax plus b, then yn equal to n derivative is minus 1 raise to n into n factorial into a raise to n divided by ax plus b raise to n plus 1. And now we will compare these two terms with 1 upon ax plus b, in this case a is 2, b is minus 1 and in second case a is 3 and b is minus 1. Therefore comparing with this formula, we will write the n derivative of this first term as yn equal to keeping this constant 2 as it is minus 1 raise to n into n factorial into a raise to n, means here a is 2, therefore 2 raise to n divided by 2x minus 1 raise to n plus 1, that is ax plus b raise to n plus 1. This first term is about the n derivative of 2 upon 2x minus 1. Now here keeping this minus sign as it is and 3 as it is and the nth derivative of 1 upon 3x minus 1 is minus 1 raise to n into n factorial into a raise to n, means 3 raise to n divided by 3x minus 1 raise to n plus 1. This is what the n derivative of the given function. Now pause the video for a while and find the n derivative of x upon x square minus 4. Pause the video and find the n derivative of this function. I hope you have completed. Now let y equal to x upon x square minus 4, that is a given function and here we have to factorize this, factorization is very simple, here we can make use of a square minus b square, you can write x square minus 4 as x square minus 2 square and factors are x minus 2 into x plus 2 using a square minus b square equal to a minus b into a plus b. Therefore, we can write x upon x square minus 4 equal to x upon x minus 2 into x plus 2 and we have to make use of partial fraction, let x upon x minus 2 into x plus 2 is equal to a upon x minus 2 plus b upon x plus 2. Therefore, multiplying both side by x minus 2 into x plus 2, you get x is equal to a into bracket x plus 2 plus b into x minus 2. Now to find a and b, you can put x equal to 2 and x equal to minus 2, putting x equal to 2 both side, we get left hand side is 2, 2 plus 2 that is 4, 4 a and second term will become 0, 2 equal to 4 a therefore, a equal to 1 by 2. Now, putting x equal to minus 2 both side, left hand side is minus 2 which is equal to here b into minus 2 minus 2 that is minus 4 b therefore, b equal to 1 by 2. Therefore, given function we can write as y equal to 1 by 2 into 1 upon x minus 2 plus 1 by 2 into 1 upon x plus 2 and therefore, y n is equal to 1 by 2 into bracket using the formula of 1 upon x plus b minus 1 raise to n into n factorial divided by x minus 2 raise to n plus 1 plus minus 1 raise to n into n factorial divided by x plus 2 raise to n plus 1 and by simplifying, you get we can take common minus 1 raise to n and n factorial that is minus 1 raise to n into n factorial divided by 2 into bracket 1 upon x minus 2 raise to n plus 1 plus 1 upon x plus 2 raise to n plus 1. Now, we will see the next example. Find the n derivative of x upon x plus 2 raise to 4. Find the n derivative of x upon x plus 2 raise to 4. Let y equal to x upon x plus 2 raise to 4. Now, see here x plus 2 is repeated 4 times and here we have to adjust the term and we have to express this in the form of 1 upon x plus b. See, we can write here x upon x plus 2 raise to 4 can be written as x plus 2 minus 2 that is here we have added to and we have subtracted to 2 without changing that is x will remain same expression will remain just in place of x we have written x plus 2 minus 2 divided by x plus 2 raise to 4. Next, which is equal to x plus 2 divided by x plus 2 raise to 4 means 1 upon x plus 2 raise to 3 minus 2 upon x plus 2 raise to 4. Now, see in the first video we have seen if y equal to 1 upon x plus b raise to m then y n is equal to m plus n minus 1 factorial into minus 1 raise to n divided by m minus 1 factorial into a raise to n divided by a x plus b raise to m plus n. Now, here we have to compare with the given expression with 1 upon a x plus b raise to m. See here a is 1 and b is 2 and here in this first case m is 3. Therefore, y n is equal to we can write m plus n minus 1 factorial here m is 3 that is 3 plus n minus 1 factorial into minus 1 raise to n divided by m minus 1 factorial means 3 minus 1 factorial into 1 upon x plus 2 raise to 3 plus n. This was the n derivative of the first term minus keeping 2 as it is and n derivative of 1 upon x plus 2 raise to 4 is 4 plus n minus 1 factorial into minus 1 raise to n divided by 4 minus 1 factorial in this above formula putting m equal to 4 into 1 upon x plus 2 raise to 4 plus n. I hope you understood just we have to we are comparing this above formula we get this one. Now, further we will simplify which is equal to therefore, y n equal to what we get minus 1 raise to n n plus 2 factorial just here we are simplifying this one 3 plus n minus 1 factorial. I we can write n plus 2 factorial that is minus 1 raise to n into n plus 2 factorial divided by 2 factorial into x plus 2 raise to n plus 3 minus 2 into minus 1 raise to n into n plus 3 factorial divided by 3 factorial into x plus 2 raise to n plus 4 and here in both term taking minus 1 raise to common which is equal to minus 1 raise to n into bracket n plus 2 factorial divided by 2 factorial is 2 into x plus 2 raise to n plus 3 minus n plus 3 factorial as it is and here 2 by 3 factorial means 1 by 3 or kept as it is that is n plus 3 factorial divided by 3 into x plus 2 raise to n plus 4. This is what the end derivative of the given function references higher engineering mathematics by Dr. B. S. Gravel. Thank you.