Nth Derivative of Exponential Function – Successive Differentiation – Engineering Mathematics 1


Hi Guys, So in this video we are going to see nth derivative of some of the functions so let’s start with the first function I am assuming that Y is equal to e raised to M X so let’s start and find out what could be the nth derivative of e raised to M X now to find out y n that is n derivative of the given function we will always start with the first derivative so my technique will be to first find out the first derivative then second derivative third derivative and then from first three answers we will predict that what could be the nth derivative so let’s start with the first derivative so first derivative is given by y1 now if I find the first derivative you all know that the derivative of e raised to M X is M into e raised to M X if I will find the second derivative it will be M square into e raised to M X similarly if I’ll try to find out nth derivative then let’s observe the pattern so in the first derivative if you’ll observe then the power of M is 1 in second derivative the power is 2 so similarly can I say that in the nth derivative the power will be n yes definitely so similarly here he raised to M X terms remains same in all the derivatives so I can say that in the nth derivative also he raised to M X will be there so the formula for y equal to e raised to M X is yn equal to M raised to n into e raised to M X so this is the nth derivative of e raised to M X let us take one more example now the second question is what is NC derivative of e raised to M now guys let’s start with the similar technique so here again we’ll find out first derivative second derivative third derivative and from these three derivatives we will guess what could be the nth derivative so I am finding the first derivative so y1 in this case will be here is 2m X into log of a into M similarly if I find the second derivative of this term then M and log a are already constant so I will write it down as it is and if I find out the derivative of a raise to M X then it will be here is 2m X into log n so log a is twice so log e square and again M that is M square similarly if I will find the third that is why three so log a square is a constant we will take it as it is M square constant we will take it as it is and then we’ll find out the derivative of a raise to M X so derivative of a raised to M X is a raised to M X into log n so this will become Q and M again so this will become M Q so hence the formula is a raised to M X log a cube and M cube so guys let’s predict the nth derivative so similarly we will get and the derivative as here is 2m x as it is why because in all the three derivatives we were getting a raise to M X term next log in here in the first derivative log it was having the power 1 in the second derivative – and the third derivative three so I can say that in antiderivative this power will be n so log a raised to n similarly for M so in y1 M raised to 1 y2 M raised to 2 y 3 M raised to 3 so I can definitely say that yn it will be M raised to n and this becomes the formula for n 3 of a raise to M X now let’s find out the next enter derivative so I am considering the function of trigonometry so let us say Y is sine of a x+ now guys be careful if I want to find out the derivative of this I will again start with the y1 y2 and y3 so here let’s see now y1 will be derivative of this is cos of a X plus B into a now let’s observe it if I’ll find out y 2 from this y1 100% for the derivative of cos I will get minus of sine again from by 2 if I find out y 3 it will be again cos and hence it will change like cos sine cos sine but guys it will be very difficult to find out y n in that case because we don’t know what is the value of n so hence to find out the antiderivative we will always convert the given function as sine only so yeah my job will be to convert this course into sine and then I will find out the further derivatives so you are if I convert the COS into sine then you all know the formula let sine of Pi by 2 plus theta is cos theta so I will say it is sine of PI by 2 plus theta that is ax plus B as it is an a was there so I have rewritten the y1 in terms of sine now let us find out why – from this y1 so again the derivative of this sign is cos of Pi by 2 plus ax plus B into a square because 1 a was there in the y1 and in the second derivative we are getting one more ring so it becomes a square but now if I will find y3 from this Casta then again I get sine so it is again mismatch as I said previously that every time we will convert the given function in terms of sine and then we’ll find out the further derivative so here I am again converting cos in terms of sine now here this PI by 2 plus ax plus B is my theta so it is my cos theta so I will rewrite it as sine of PI by 2 plus theta because sine of PI by 2 plus theta is cos theta so this will become sine of PI by 2 plus theta now theta is this term that is PI by 2 plus ax plus B and a square as it is so now guys Here I am getting PI by 2 2 times so I will rewrite this as sine of 2 times pi by 2 plus ax plus B into a square this is my waifu now if i try to predict what could be the wire then it’s very easy see here in why I was just having angle as ax plus B in y1 I got angle as ax plus B plus Phi by 2 that is PI by 2 is only once in Y – I am getting angle as ax plus B and PI by 2 twice so it means the PI by 2 terms is increasing every time so can I say when I will find the antiderivative I will get PI by 2 n times so hence this becomes sine of a X plus B plus n PI by 2 similarly if you observe the constant a in the first derivative y1 it was just a raised one in y2 it was a square and similarly in Y n we will get a raise to n and this becomes our formula for the nth derivative of sine of a X plus B