# Interactive video lesson plan for: unit 4 test application of derivatives #7 Linearization

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Good day students welcome to mathgotserved.com in this clip we're going to be going over problem number seven of all you unique for tests on the application of derivatives okay and this am always going to be focusing on the Linearization for problem number seven we are to find the linearization of X of F of X equals one over one plus ten X at X eighty equals zero okay I now let's read of the formula for immunization first and then proceed to find you irritation of this function linearization of the function L of X is equal to F of a plus F prime of a times X minus a right so this formula the same formula asked the point slope form of the equation of the line that you learn in algebra one into so we have why equals why one plus and X minus X one right so the sinking this is a much more advanced notation for the equation of the tangent line I have to get ready for the more complex topics in calculus to and three okay so we going to be speaking to the more advanced notation this is about and this notation is elementary okay is limited in its application isolate go ahead and do this so if you think back to other cool one to if you are want to write equation of the line you needed points on the slope of exactly what we need here okay so in this format using this advanced notation the points is a, F of a and the slope and is basically F prime of a right you can see why one X one and in F prime of a right so let's go ahead and find the components that we need first we find a a is given a is zero that is the x-coordinate of the points that we find the linearization of the curve at okay what is the y-coordinate X so we need to find F of a which is F of zero since a is zero we going to about with this function at a equals zero so we have one over one plus ten of zero okay so is going to have one over one plus ten zero is what zero zero if you remember you think about your tangent curve sketch that real quick is ten curve look something like this since concave down and then concave up and concave down so at zero, zero zero or he can think about time zero as sine zero over cosine zero and we know that Cosine zero is zero and cosine - zero is zero and cosine zero is one goes your one is just a okay so we have one over one zero equals one so that is yours y-coordinate what do we need next to the F prime of a okay enough find F prime of the we need to do two things first we have to find the derivative F prime of X right and then secondly we need to about with the derivative at a which is basically going to be F prime of zero are isolate two-step process now let's go ahead and start by finding the derivative let's write a function F of X I want to find the derivative of this function F of X is equal to one plus ten X okay so to find the derivative take care F prime of X is one over one plus ten X prime so I can use the quotient rule here but if you notice the numerator is a constant so I can simply express this using the reciprocal properties of exponents I simply use the chain rule so this can be written as one plus X raised to the negative one so looking for the derivative of this composite function right on using chain rule you can use the chain role but that's too much I select hundred derivative using the chain rule the derivative of the outside function on this case the outer function is X the negative one is derivative is negative X to the negative to the inner function is one plus ten X and its derivative is zero plus can square X okay right so let's apply that here differentiate the outside if I waited at the inside so we have negative one times one is tan X reason negative to times the derivative of the inner function is going to be zero plus can square X right now that F prime of X we do the first part one YouTube I waited at a equals zero okay so have one negative one times one plus ten zero this entire the reason is to times just zero secant's square of zero okay isolate and about with this for that have negative one remember tan of zero we talked about that relates zero so we have one for zero is one raised to negative to secant square zero is secant zero square okay secant zero this is a reciprocal treat function you can written as one over cosine zero and we know from the graph of the cosine curve at cosine of zero is one okay so we have one over one which is one so secant of zero is one so we have this times one square since we square is secant beta simplify this you have negative one times one times one okay the reciprocal of one is one when you square X just get one right so now we have our slope F prime of zero is negative one so we happily components we need to write equation of our of the linear it to write a linearization okay remember the linearization is just equation of the tangent line right so let me write a formula again so that we can apply our results to this

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