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Good day students welcome to mathgotserved.com in this clip we are going to be going over how to apply one of the differentiation techniques which is the chain rule we are going to apply it in finding the derivative of composite functions dont forget to visit our website at mathgotserved.com for access to a wide variety of math tutorials ranging from algebra to calculus alright before we get started lets go over the chain rule real quick the formula that is going to guide our differentiation process so the chain rule the chain rule is basically applicable to composite functions okay lets just take a look at the composition of two functions so lets say we have a composite function y=f(g(x)) I am going to color code it so we can keep track of what is going on so lets say we have this composite function y composed with g(x) the derivative y prime is going to be the derivative of this composite function [f(g(x))]' the entire composite function prime the chain rule is that the derivative of this composite function is f'(g(x)) multiplied by g'(x). alright so another name for the chain rule is known as the outside inside rule basically you have a function on the outside and then another function on the inside so which function is on the outside and which function is on the inside? If you take a look at the original situation that we have right here f is on the outside this is the function on the outside and so lets call this the outside function and g is the function on the inside okay so we are composing f with g so the derivative basically tells you that you are going to find the derivative of the outside function the derivative of the outside function evaluated at the inside function okay so you differentiate the outside function and then you plug in the inner function lets call it the function on the inside and then umm you multiply that by the derivative of the inside function okay inside function prime so outside function prime of the inside function times the inside function prime will give us the derivative now this notation looks very confusing a lot of students confuse the functions the outer and the inner functions so in order to keep it clean and avoid confusion and clutter most textbooks make use of a substitution called the u substitution okay so to keep our work nice and pretty we are going to say let the inner function g(x) be u so let u equal g(x) alright so that is going to give us the formula we are going to be using today which is f' of the inner function instead of g(x) just to make it look good we are going to call it u times u' okay time u prime so the derivative of the outside evaluated at the inner function times the derivative of the inside function alright this is equivalent to this is known as the lagrange notation the leibniz notation you see it in some books if you have d/dx of a composite function f(u) okay you have a composite function like this this is the same thing as lets just call it the derivative of the compsite function y so d/dx of y. y is a composite function dy/dx umm it is going to be written as dy/du times du/dx now this is exactly thesame thing as what we have here dy/du is the derivative of the outside function evaluated at the inner function and du/dx is the derivative of the inner function okay so these two formulas are exactly thesame thing okay. alright so with that in mind let us go ahead and try two examples now what we are going to be doing today we are going to be taking a look at some AP calculus type questions the kind of questions you will see umm in the AP calculus exam so lets take a look at question 1 find y' where y is the composite function y=(x^3-cosx)^5 we are going to go step by step with this problem so that everybody understands it first thing we are going to do let me rewrite my function as a composition of two functions so let y=f(g(x)) okay now so f(g(x))) which is equals to x^3 lets color code our functions so the inner function is x^3-cosx and that entire quantity raised to the fifth power alright so before we start making our u substitution and applying this formula right here, lets go ahead and identify our outer and inner functions okay so that is the formula that I am going to be using so my outer outside function we can call it outer outside is what is f(x) its the fifth power polynomial so x^5 that is the outer function the outside function . so the inside function do you see what the inside function is? lets color code that also the outside function is x^5 the inside function is inside function lets call that g is umm x^3-cosx okay so the inside function is x^3-cos x now if you compose f and g hopefully you can see that this will be the result. If you compose f and g you evaluate f(x) at x^3-cos x namely plug this into the x you have x^3-cosx raised to the fifth power okay. so now we know our outside and inside function to make this work pretty and easy on the eyes we use of the u substitution
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