 Hello, Namaste. Myself, MS Bhashargaon, Assistant Professor, Department of Humane Design Sciences, Walsh and NISTOP Technology, Solapur. In this video, we are going to discuss Successive Differences in Part 3, Learning Outcome. At the end of this session, students will be able to find nth derivative of algebraic and logarithm functions. Now first of all, we will see the example on algebraic function that we have already seen the formulas of algebraic function. Find nth derivative of x square upon 2x square plus 7x plus 6. Now here, look at here, here the degree of the numerator is 2 and denominator is also 2. Therefore, we cannot go for partial fraction directly. Therefore, here first of all, we have to divide numerator by denominator because we have to express the function in the form of 1 upon ax plus b and for that, we have to go for the partial fraction. But here, the degree of the numerator and denominator is same. Therefore, before going for the partial fraction, we have to divide numerator by denominator. That is, we are dividing x square by 2x square plus 7x plus 6. That is here, to get the coefficient of x square is 1, we have to multiply 2x square plus 7x plus 6 by 1 by 2. That what we get? x square plus 7 by 2x plus 6 by 2 means plus 3. That is subtracting, we get x square x square, we will get cancel minus 7 by 2x and minus 3. You know how to divide the one polynomial by other polynomial. And now, see we can write, first of all, now we will here, factor it 2x square plus 7x plus 6. 2x square plus 7x plus 6, we can write 2x square plus 4x plus 3x plus 6. Here, we have adjusted the middle term. And in first two terms, we can take 2x common, that 2x into bracket x will remain x plus 2. And here, in the second term, we can take 3 common, that is 3 into bracket x plus 2. Taking x plus 2 common, we will get 2x plus 3. Therefore, factors of 2x square plus 7x plus 6 are x plus 2 into 2x plus 3. Therefore, we can write the given function, x square upon 2x square plus 7x plus 6 as here by using co-centered remainder theorem, we can write y equal to 1 by 2. This taking minus sign common minus of 7 by 2x plus 3 divided by 2x square plus 7x plus 6 means x plus 2 into 2x plus 3. Now, we can go for partial fraction. Let minus of 7x by 2 plus 3 divided by x plus 2 into 2x plus 3 equal to a upon x plus 2 plus b upon 2x plus 3. In multiplying both side by x plus 2 into 2x plus 3, you get minus of 7x by 2 plus 3, which is equal to a into 2x plus 3 plus b into x plus 2. And now, we have to find the values of a and b by substituting the suitable values of x, that is, first we have to put x equal to minus 2 both side. If you put x equal to minus 2 both side, you get left hand side value will be 4. And here what we get minus a therefore, a equal to minus 4. Now, similarly putting x equal to minus 3 by 2 both side, we get minus of x equal to minus 3 by 2 means minus 21 by 4 plus 3 which is equal to b into minus 3 by 2 plus 2. That is simplification of this left hand side is 9 by 4 which is equal to and this right hand side simplification is 1 by 2 that is 1 by 2 b therefore, b equal to 9 by 2. Now, substituting the values of a and b, you get the given function y equal to 1 by 2 minus 4 by x plus 2 plus 9 by 2 into 1 upon 2x plus 3. Therefore, now see here, we have expect the given function in the form of 1 upon ax plus b here first term is constant. Now, we have to we know that in last video we have seen if y equal to 1 upon ax plus b then yn equal to minus 1 raise to n into n factorial into a raise to n upon ax plus b raise to n plus 1. Now, first thing is here, the derivative of 1 by 2 is 0 because derivative of constant term is 0 minus 4 as it is and comparing 1 by x plus 2 with 1 upon ax plus b here a is 1. Therefore, 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 plus 9 by 2 as it is and n derivative of 1 upon 2x plus 3 is minus 1 raise to n into n factorial into 2 raise to n divided by 2x plus 3 raise to n plus 1 comparing with here a and b here a is 2 and b is 3. This is what the n derivative of the given algebraic function that is by simplifying further simplifying you get y n equal to taking minus 1 raise to n common that is minus 1 raise to n into n factorial and writing the first positive term that is 9 into 2 raise to n minus 1 that 2 raise to n by 2 means 2 raise to n minus 1 divided by 2x plus 3 raise to n plus 1 minus 4 divided by x plus 2 raise to n plus 1. This is what the simplified form of n derivative of given function. See we have seen the sufficient number of examples on algebraic function now we will go for the next type of function that is n derivative of log of ax plus b that is n derivative of logarithm function log of ax plus b. If y equal to log of ax plus b then y n is equal to minus 1 raise to n minus 1 into n minus 1 factorial into a raise to n divided by ax plus b raise to n. Now, we will see the proof of this we have when y equal to log of ax plus b let us call this as equation number 1. Differentiating 1 with respect to x we get that you know the derivative of log of f of x that is what 1 upon f of x into a appetite of x therefore, derivative of log of ax plus b is 1 upon ax plus b into derivative of ax plus b is a therefore, first derivative of log of ax plus b is a upon ax plus b. And we know if y equal to 1 upon ax plus b then y n equal to minus 1 raise to n into n factorial into a raise to n divided by ax plus b raise to n plus 1 that is we have already seen the proof of this. Now, differentiating this n minus 1 times with respect to x difference because already here we have calculated the first derivative and to get the n derivative from first derivative differentiating n times with respect to x we get y n equal to with help of this or to get minus 1 raise to n minus 1 in this one you are replacing minus n by n minus 1 that is minus 1 raise to n minus 1. Then n minus 1 factorial replacing n by n minus 1 here we get n minus 1 factorial and a raise to n minus 1 and already here a is there that is what we get a raise to n divided by ax plus b raise to n in this one also replacing n by n minus 1 what we get ax plus b raise to n. Therefore, when y is equal to log of ax plus b then y n is minus 1 raise to n minus 1 into n minus 1 factorial into a raise to n divided by ax plus b raise to n. Now pause the video for a while and find the n derivative of log of 5x minus 3 whole square. I hope you have completed let y equal to log of 5x minus 3 whole square and which we can write this as a twice log of 5x minus 3 we know the log of a raise to b as a b log a by using the property we can write twice log of 5x minus 3 and now you comparing with log of ax plus b here a is 5 and b is minus 3 and just now you have seen if y equal to log of ax plus b then y n equal to minus 1 raise to n minus 1 into n minus 1 factorial into a raise to n divided by ax plus b raise to n. Now here a is 5 and b is minus 3 therefore, y n is equal to keeping to these two as it is minus 1 raise to n minus 1 n minus 1 factorial into a raise to n means here a is 5 that is 5 raise to n divided by 5x minus 3 raise to n this is what the n derivative of log of 5x minus 3 whole square. Now we will see the example find the n derivative of log of 4x square minus 1 let y equal to log of 4x square minus 1 here we have to express this logarithm function in the form of log of ax plus b which is equal to we can write log of 4x square can be written as 2x whole square minus 1 square which is equal to log of 2x minus 1 into 2x plus 1 here using the formula of a square minus b square as a a minus b into a plus b that is 2x square minus 1 square we can write 2x minus 1 into 2x plus 1 and now using the property of logarithm log of a into b as a log a plus log b that is equal to log of 2x minus 1 plus log of 2x plus 1. Now we have the formula if y equal to log of ax plus b then yn equal to minus 1 raise to n minus 1 into n minus 1 factorial into a raise to n divided by ax plus b raise to n now we can apply this formula here. Now here if you compare with log of ax plus b in both the cases a is 2 therefore yn is equal to minus 1 raise to n minus 1 into n minus 1 factorial into 2 raise to n divided by 2x minus 1 raise to n plus minus 1 raise to n minus 1 into n minus 1 factorial and 2 raise to n divided by 2x plus 1 raise to n this is what the n derivative of the second term. Now taking the common terms here we can take minus 1 raise to n minus 1 common n minus 1 factorial common and 2 raise to n also common taking the common then in the first term we get 1 upon 2x minus 1 raise to n plus 1 upon 2x plus 1 raise to n this is what the n derivative of log of 4x square minus 1 references higher engineering mathematics by Dr. B. S. Gravel. Thank you.