75–86. Logarithmic differentiation Use logarithmic differentia...

Question

75–86. Logarithmic differentiation Use logarithmic differentiation to evaluate f'(x).

f(x) = x^In x

Accepted Answer

Welcome back, everyone. In this problem, we want to use logarithmic differentiation to evaluate the derivative of the function G of X equals X E X. A says it's X E X minus 1 multiplied by EX multiplied by XLN X + 1. B E X multiplied by X E X minus 1. C X E X minus 1 multiplied by EX multiplied by XLN X minus 1, and the D X E X multiplied by the natural log LN of E to the X minus 1. Now, if we're going to evaluate this derivative using logarithmic differentiation, we have to ask ourselves, how can we introduce logs into our function? Well, let's let Y be equal to G of X. Then our function is going to become Y equals X E X, OK? Now if we take the natural logarithm of both sides. Then we're going to get the natural log of Y to be equal to the natural log of X to the E to the X. And that's pretty convenient because you recall that based on the properties of logs, if we are finding the log of a term race or poor, we can bring down the power. In other words, the natural log of A to the X is the same as XL and A. So if we apply that idea here. Then we can say that LN Y is equal to EX multiplied by LNX. Now we can use logarithmic differentiation to evaluate our derivative, which has now been written in a simpler way. Notice that on the left-hand side, we have the natural log of Y, so we'll have to differentiate implicitly, OK? Now by differentiating implicitly. OK. If we think about it, On the left hand side, the derivative of LNY, uh, well, let's write it here. The derivative of, we're gonna have the derivative of LNY multiplied by the derivative of E to the XLNX. And again, back to our left hand side, the derivative of LNY is gonna be 1 divided by Y multiplied by Y. Now, what about what's going on on our right-hand side? Well, here we notice that E to the XLNX is our product. So that means we can differentiate it using the product rule of differentiation. And by the product tool. Let me just write a note to the right here. By the product rule, it basically says that if Y equals UV where U and V are differentiable functions, then the derivative of Y is going to be equal to U V plus UV. So we're going to differentiate both terms in our product and plug them into the product rule. So from our product here on the on the right hand side, if we let you be equal to E X and V equals L and X, then the derivative of U is E X and the derivative of V is going to be 1 divided by X. Therefore, if we apply the product rule, OK, then on our right hand side, we're going to have. U V, OK, so that's E to the XLN X. Plus UV e to the X multiplied by 1 divided by X. OK, so that's the derivative on our right hand side. Notice that since they both have a factor of e to the X we can factor it out so that's e X multiplied by LNX plus 1 divided by X and know if we combine both of these terms, OK. Into one fraction here, OK? Then we're gonna have E to the X multiplied by LNX plus, sorry, XLN X. Plus 1 divided by X, OK. Now we could multiply both sides by Y as well, OK, to get, because remember we're solving for the derivative of Y, OK? So that's gonna be Y equals Y E to the X, multiplied by XL and X + 1 divided by X. And remember earlier we defined Y as geoff X. So if we replace it, OK, with a function G of X, that's gonna be X to the E to the X multiplied by E to the X. Multiplied by X and X + 1 are divided by X. No, notice here if we, if we go ahead here and apply what we know about uh the the powers, the rule of powers, yeah. Notice this isn't denominator, it's the same base so we could take a step further and rewrite this as X to the E X minus 1 to account for the X and the denominator multiplied by E to the X multiplied by XLN X + 1. Thus, this is the derivative of Y, which is the derivative of G of X. Therefore, when we look back at our answer choices, that tells us that A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

75–86. Logarithmic differentiation Use logarithmic differentiation to evaluate f'(x).
f(x) = x^In x

Key Concepts

Logarithmic Differentiation

Chain Rule

Exponential and Logarithmic Functions

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