# Higher Order Derivatives

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**Higher Order Derivatives**. Following along are instructions in the video below:

in this video were gonna focus on finding higher-order derivatives so lets say if we have the function f of X and its equal to three X to the fifth power plus two x cubed minus 6x plus four and in this problem go ahead and determine the second derivative of the function well we need to find the first derivative to begin so the derivative of X to the fifth using the power rule is 5x to the fourth and the derivative of x cubed is three x squared and the derivative of X is one and a derivative of a constant is zero so this is going to give us 15 X to the fourth 2 times 3 is 6 and so this is the first derivative now we need to differentiate this expression one more time to get the second derivative so if we differentiate X to the fourth thats going to be 4x cubed and the derivative of x squared is 2x so the second derivative is going to be, 15 times four which is 60 and six times two is, 12 so its going to be 60 X cubed plus 12x now lets try another problem lets say that H of X is x squared cosine X go ahead and determine the second derivative of this function now for this problem we need to use the product rule the derivative of F times G is going to read the derivative of the first part F prime plus 4 times the second part plus the first part times the derivative of the second part so we could say that F is x squared and G is cosine so if f is x squared f prime has to be 2x and if G is cosine G prime the derivative of cosine is going to be negative sine so using this formula H prime of X is going to equal the derivative of the first part which is 2x times the second part thats cosine plus the first part which is just f 4x squared times the derivative of the second part which is negative sine X now we could factor the expression or we could leave it like this right now its best to leave it the way it is so we have 2x cosine X minus x squared sine X now lets go ahead and find a second derivative so were gonna have to use the product rule twice for this term and for that one so for the first term lets separate it into two functions so this is the first part thats f and heres the second part G so H double prime is going to be the derivative of the first part the derivative of 2x is two and

then times second part which were going to leave the same cosine plus were going to keep the first part the same thats F times G prime the derivative of the second part the derivative of cosine is negative sine now lets focus on x squared sine X so lets say this is f and this is going to be G so the derivative of the first part x squared is 2x times the second and then plus the first part x squared times the derivative of the second part which is cosine so now lets rewrite everything so we have two cosine – 2x sine X and dont forget to distribute this negative sign so – another 2x sine X minus x squared cosine X so it looks like we can combine like terms so the final answer is going to be 2 cosine X – 4 X sine X minus x squared cosine X so thats H double Prime now lets say if f of X is equal to the square root of x and we need to find a third derivative F triple prime of X how can we do so so go ahead and try that problem now first we need to rewrite the expression so the square root of x is the same as X to the 1/2 and lets use the power rule so keep in mind the power rule is n X raised to the N minus 1 so its gonna be 1/2 X raised to the 1/2 – 1 now 1/2 minus 1 is basically 1/2 – 2 over 2 which is a negative 1 over 2 so this is the first derivative now I wouldnt recommend rewriting it and simplifying it yet until you find the third derivative so lets go ahead and find a second derivative while its in this form so the derivative of x to the negative 1/2 thats going to be negative 1/2 X and then negative 1/2 – 1 now 1/2 times negative 1/2 thats a negative 1/4 a negative 1/2 – 1 think of negative 1/2 minus 2 over 2 negative 1 minus 2 is negative 3 so this is going to be X raised to the negative 3 over 2 now lets find the third derivative so using the constant multiple rules can be negative 1/4 times the derivative of that expression so thats negative 3 over 2 X raised to the negative 3 of a 2-1 now negative 1 times negative 3 thats gonna be positive 3 and then 4 times 2 is 8 now negative 3 over 2 minus 1 or negative 3 over 2 minus 2 over 2 thats negative 5 over 2 so this is the answer all one needs to do is simplify or rewrite

it so right now the third derivative is equal to 3 X to the negative 5 over 2 divided by 8 now to make the negative exponent positive we need to move the variable to the bottom so its going to be 3 over 8 X raised to the positive 5 over 2 now we can convert this into its radical form so its 3 over 8 square root X to the fifth power now we could still simplify this expression further if we want to X to the fifth is X to the fourth times X and the square root of x to the fourth keep in mind the index numbers 2 so its 4 over 2 which is 2 so its gonna be x squared thats the square root of x to the fourth now we could still rationalize the denominator if you want to so if we multiply the top and the bottom by the square root of x were gonna have 3 square root x over 8x squared times the square root of x squared which is X and x squared times X is X cubed so you can write your answer as 3 square root X divided by 8x cubed if you want to fully simplify it and rationalize the denominator at the same time now lets work on one more problem lets say if youre given the second derivative and its 5 over x squared go ahead and determine the fourth derivative of the function now first we need to rewrite the expression so Im going to move this to the top so this is the same as 5x to the negative 2 so we already have the second derivative lets find a third so we need to find the derivative of 5x to negative 2 so its gonna be 5 times the derivative of x to the negative 2 which is negative 2 X to the negative 2 minus 1 which is negative 3 and so 5 times negative 2 thats a negative 10 so the third derivative is negative 10 X to the negative third so now we need to find a fourth derivative so its gonna be negative 10 times the derivative of X to the negative 3rd which is negative 3 X to the negative 3 minus 1 which is negative 4 a negative 10 times negative 3 is 30 so we have 30 X to the negative 4 which we can rewrite as 30 divided by X to the 4th power and so if youre given the second derivative you need to differentiate it two more times to get to the fourth derivative and so thats just the basis of higher order derivative problems you just got to find a second third or fourth derivative

tags:

higher order derivatives, second derivatives, calculus, derivatives, how to find the second derivative of a function, introduction

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