# Video: MATH-DIFF-INT-2018-S1-Q10

If π(π₯) = π₯(π β ln(π₯)) such that π is a constant and the curve of the function has a critical point at π₯ = π, find π.

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### Video Transcript

If π of π₯ is equal to π₯ lots of π minus ln of π₯ such that π is a constant and the curve of the function has a critical point at π₯ equals π, find π.

First, letβs remember what a critical point is. A critical point is a point, such as π₯ is equal to π, such that the differential of π at π, which we can also call π dash of π, is equal to zero. So letβs start answering this question by differentiating π of π₯ with respect to π₯. And this is also called π dash of π₯. And itβs equal to d by dπ₯ of π₯ times by π minus ln of π₯. In order to differentiate this, weβll need to use the product rule. The product rule tells us that if we have some function π¦, which is equal to a product such as π’ multiplied by π£, then when we differentiate π¦, weβll get dπ¦ by dπ₯ is equal to π’ dπ£ by dπ₯ plus π£ dπ’ by dπ₯.

Now, in our case, we can let π₯ be π’ and let π minus ln of π₯ be π£, then we obtain that this differential is equal to π₯ times by d by dπ₯ of π minus ln of π₯ plus π minus ln of π₯ times d by dπ₯ of π₯. And, immediately, we can evaluate d by dπ₯ of π₯, since this is just equal to one. In order to differentiate π minus ln of π₯ with respect to π₯, weβll need to use a rule. And this rule tells us that d by dπ₯ of ln of π₯ is equal to one over π₯. Now, in order to answer this question, you do not need to know how to derive this differential. But for those of you who are interested, I shall write down the derivation of this differential. And you can pause the video and read through how itβs done.

So this is how you can prove that d by dπ₯ of ln of π₯ is equal to one over π₯. However, itβs not necessary to know this in order to answer the question. All you need to remember is that the differential of the natural logarithm of π₯ is equal to one over π₯. So now we can use this in order to resolve the differential of π minus ln of π₯. Now, weβre told in the question that π is a constant. And any constant will differentiate to zero. So the π will go to zero, and then we have minus ln of π₯. And since ln of π₯ differentiates to one over π₯, negative ln of π₯ will differentiate to negative one over π₯. And then, we need to add on the π minus ln of π₯ times by one.

Now, we can simplify this to give us that π dash of π₯ is equal to negative one plus π minus ln of π₯. Now that we found π dash of π₯, we need to use the fact that this function has a critical point at π₯ equals π. If we remember back to the definition, if a critical point of a function is at π₯ equals π, then this means that π dash of π is equal to zero. Since our critical point is at π, we need to substitute π₯ equals π into this equation and set it equal to zero.

This gives us that π dash of π, which is equal to negative one plus π minus ln of π, is equal to zero. Now, since ln is the natural logarithm or log to base π, ln of π is simply one. And, therefore, our equation becomes π minus two is equal to zero. This can be arranged for π to give us that π is equal to two. And this is our solution.