# Interactive video lesson plan for: 2.5 Higher Order Derivatives f f" ap calc calculus leibniz lagrange

#### Activity overview:

Good day students welcome to mathgotserved.com in this clip we are going to be going over how to computes higher-order derivatives okay so before we get started let's take a look at the notation for expressed in higher-order derivatives right we going to be looking at two times so let's of expressing higher-order derivatives the going to look at all the Lagrange notation and the Leibniz notation okay so for the function function why going to look at the first and second in their derivatives are isolate is seeking chart here to keep our work organize going to look at the Lagrange notation in the Leibniz notation okay right so what if you wanted to find the firsts derivative right to the first derivative using the Lagrange notation is why prime in the Leibniz notation is the why the X here the one here anyone here but or the first derivative you don't bother putting that number there are okay now how about the second derivative for the second derivative forty Lagrange notation is why double prime and then for the Leibniz notation is the square why the X square okay over infinity pattern here to how to generate all the derivative expression using the two notation okay so formula right notation the number of primes indicates the derivative bets you taking order number times you to differentiate the original function to arrive that is our derivative Leibniz you look at the power of the constant is the order what the BX powers race to that's what tells you what the derivative is okay so let's go ahead and make a generalization so what of you looking at what you are looking for that nth derivative so the Lagrange notation for the X derivative is why parenthesis and Y to the end our axis (and effort to live this is the and Y the X the X okay so this is just a real quick overview of the notation that were going to be using to guide our problem-solving process so let's go ahead and try out some examples right so for problem number one in why is equal to sine X applying why double prime okay is for that again why double prime now calling times I we expected to differentiate this function to find why double prime this is the Lagrange notation for derivatives number prime tells you probably times you to differentiate the original function so since we have to product here we're going to be differentiate his twice okay so we have why equals sine X over differentiate first why prime why prime is going to be the derivative of cosine sorry derivative of sine which is cosine X now let's differentiate again the derivative of why prime is why double prime which is equal to the derivative of cosine X which is negative sine X okay so there goes your final answer right let's take a look at question number two so for question two we are to find the potential one so the question and show points of inflection four the function why equals three X to the third plus two X square okay so what is this problem ask interest to do what is involved in finding the potential point of inflection so you have a point of inflection points of inflection at a particular point X equals a is you have to conditions okay condition number one the second derivative of the function at that point is equal to zero and on their the concavity changes okay so why change is concavity what does that mean wine changing concavity means that why changes from concave up to concave down at a or changes from concave down to compute a also okay so one of the two conditions another is a changing concavity and the second derivative is zero then you have a point of inflection okay so it doesn't people potential points here so find the second derivative and said it equal to zero is something for X will help us find the potential points of inflection okay in other to the ensure that is a sure that the are in fact once of inflection we have to do this concavity test at that point okay so let's go ahead and solve the problems what we doing here be degrees one is solve the equation one assault why double prime of X equals zero that's the goal is going to do that we know that why equals three X to the third plus two X square the first derivative why prime is equal to nine X square plus four X the second derivative why prime which houses find the potential point of inflection all is going to be eighteen X plus four K not to find the X values where you have the potential point of inflection we said the second derivative to zero so within a set equal X is for two zero and also this would have X equals negative four over eighteen which reduces to negative to over nine okay so our potential points of inflection is at X equals negative two over nine right so now let's you determine if it so point of inflection we just look at I what happened before and after to see there is a changing concavity here and there is an is the points of inflection okay positive take a look at the last problem

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