How to differentiate ln (e^2x/( e^x-1) Derivative Differentiation Collection Chain Rule CR9 AP Calc

maths gotserved
Published at : 24 Dec 2020
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The following video will go over the derivative collection example, in finding the natural logarithm of (e^(2 x))/(e^(2 x)-1). The fist thing that must be done when the problem is displayed is to decompose the function into its components. The way to know how to decompose the function or if to decompose the function is to see if the equation has a composite function, in this case being the natural logarithmic function which is being composed with the quotient of two exponential functions. Once the function is decomposed the chain rule will be applied, the first step to the chain rule is to separate the function into f(x) and g(x). In the case of this equation, f(x) is the natural logarithm of x (ln x) and g(x) is the two exponential functions which is [(e^(2 x))/(e^(2 x)-1)]. Remembering the differentiation rules for natural logarithms we know that when (ln x) is differentiated then the answer is (1/x), therefore, based on differentiation, f(x)`= (1/x). Moving on to differentiating g(x) we know that because two exponential functions are being divided and the derivative of them is needed to be found the quotient rule will be used. The quotient rule, if it can be recalled is [u/v]`=(v(u)`-u(v)`)/(v)^2. (e^(2 x)) can be labeled as v and (e^(2 x)-1) can be labeled as u. If the two equations were to be plugged in correctly to the quotient rule formula then it would be set up as follows: ((e^(2 x)-1)[(e^2 x)]`-(e^(2 x))[e^(2 x)-1]`)/(e^(2 x)-1)^2, the derivative of e^(2 x) is (e^2 x) times 2 and the derivative of (e^(2 x)-1) is also (e^2 x) times 2. When the proper factorization is done to the equation, you will be left with (2 x^(2 x)(e^(2 x)-1-e^(2 x))/(e^(2 x)-1)^2, and once simplification of the equation is done the final result is ((-2 x^(2 x))/(e^(2 x)-1)^2). Once the right rules have been applied the chain rule can continue to be put into use, since the differentiation of f(x) is (1/x) and the differentiation of (e^(2 x))/(e^(2 x)-1) is ((-2 x^(2 x))/(e^(2 x)-1)^2) you must plug in the original g(x) into the x of the differentiated f(x) in this case being (1/x) and multiply by the differentiated function of g(x) which in this case is ((-2 x^(2 x))/(e^(2 x)-1)^2). The problem can be set up as y`=((e^(2 x)-1)/(e^(2 x))) multiplied by ((-2 e^(2 x))/(e^(2 x)-1)^2). the equation can be broken down in order to make simplification of it easier for the problem solver, the denominator of the second function the the equation is being multiplied by can be broken down into (e^(2 x)-1)(e^(2 x)-1) instead of having it stay as (e^(2 x)-1)^2). Once the equations are multiplied straight across the equation will end up being y`=(-2)/(e^(2 x)-1) because in the step before (e^(2 x)-1) can be canceled out on the numerator and denominator of the function and (e^(2 x)) can also be canceled out on the numerator and denominator of the function which is why the final answer for the differentiation of the natural logarithm of (e^(2 x))/(e^(2 x)-1) is y`=(-2)/(e^(2 x)-1).