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A bit of a classic implicit differentiation problem

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is the problem y is equal to x to the x.

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And then to find out what the derivative of y

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is with respect to x.

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And people look at that, oh you know, I don't have just a

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constant exponent here, so I can't just use the power

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rules, how do you do it.

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And the trick here is really just to take the natural log of

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both sides of this equation.

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And this is going to build up to what we're going to

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do later in this video.

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So If you take the natural log on both sides of this equation,

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you get the natural log of y is equal to the natural

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log of x to the x.

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Now our power rules, or I guess our natural log rules, say

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look, if I'm taking the natural log of something to the

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something, this is equivalent to, I can rewrite the natural

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log of x to the x as being equal to x times the

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natural log of x.

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So let me rewrite everything again.

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If I take the natural log of both sides of that equation, I

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get the natural log of y is equal to x times the

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natural log of x.

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And now we can take the derivative of both sides of

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this with respect to x.

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So the derivative with respect to x of that, and then

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the derivative with respect to x of that.

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Now we're going to apply a little bit of the chain rule.

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So the chain rule.

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What's the derivative of this with respect to x?

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What's the derivative of our inner expression

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with respect to x?

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It's a little implicit differentiation, so it's dy

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with respect to x times the derivative of this whole

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thing with respect to this inner function.

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So the derivative of natural log of x is 1/x.

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So the derivative of natural log of y with

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respect to y is 1/y.

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So times 1/y.

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And the derivative of this-- this is just the product rule,

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and I'll arbitrarily switch colors here-- is the derivative

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of the first term, which is 1, times the second term, so times

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the natural log of x plus the derivative of the second term,

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which is 1/x times the first term.

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So times x.

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And so we get dy/dx times 1/y is equal to natural log of x

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plus-- this just turns out to be 1-- x divided by x, and

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then you multiply both sides of this by y.

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You get dy/dx is equal to y times the natural

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log of x plus 1.

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And if you don't like this y sitting here, you could just

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make the substitution.

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y is equal to x to the x.

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So you could say that the derivative of y with respect to

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x is equal to x to the x times the natural log of x plus 1.

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And that's a fun problem, and this is often kind of given as

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a trick problem, or sometimes even a bonus problem if people

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don't know to take the natural log of both sides of that.

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But I was given an even more difficult problem, and

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that's what we're going to tackle in this.

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But it's good to see this problem done first because it

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gives us the basic tools.

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So the more difficult problem we're going to

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deal with is this one.

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Let me write it down.

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So the problem is y is equal to x to the-- and here's the

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twist-- x to the x to the x.

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And we want to find out dy/dx.

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We want to find out the derivative of y

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with respect to x.

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So to solve this problem we essentially use the same tools.

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We use the natural log to essentially breakdown this

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exponent and get it into terms that we can deal with.

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So we can use the product rule.

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So let's take the natural log of both sides of this equation

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like we did last time.

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You get the natural log of y is equal to the natural log

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of x to the x to the x.

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And this is just the exponent on this.

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So we can rewrite this as x to the x times the natural log

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times the natural log of x.

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So now our expression our equation is simplified to the

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natural log of y is equal to x to the x times the

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natural log of x.

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But we still have this nasty x to the x here.

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We know no easy way to take the derivative there, although I've

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actually just shown you what the derivative of this is, so

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we could actually just apply it right now.

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I was going to take the natural log again and it would turn

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into this big, messy, confusing thing but I realized that

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earlier in this video I just solved for what the

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derivative of x to the x is.

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It's this thing right here.

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It's this crazy expression right here.

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So we just have to remember that and then apply and

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then do our problem.

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So let's do our problem.

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And if we hadn't solved this ahead of time, it was kind of

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an unexpected benefit of doing the simpler version of the

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problem, you could just keep taking the natural log of this,

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but it'll just get a little bit messier.

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But since we already know what the derivative of x to the

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x is, let's just apply it.

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So we're going to take the derivative of both

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sides of the equation.

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Derivative of this is equal to the derivative of this.

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We'll ignore this for now.

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Derivative of this with respect to x is the derivative of

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the natural log of y with respect to y.

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So that's 1/y times the derivative of y

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with respect to x.

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That's just the chain rule.

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We learned that in implicit differentiation.

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And so this is equal to the derivative of the first term

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times the second term, and I'm going to write it out here just

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because I don't want to skip steps and confuse people.

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So this is equal to the derivative with respect to x of

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x to the x times the natural log of x plus the derivative

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with respect to x of the natural log of

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x times x to the x.

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So let's focus on the right hand side of this equation.

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What is the derivative of x to the x with respect to x?

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Well we just solved that problem right here.

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It's x to the x natural log of x plus 1.

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So this piece right there-- I already forgot what it

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was-- it was x to the x natural log of x plus 1.

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That is x to the x times the natural log of x plus 1.

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And then we're going to multiply that times

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the natural log of x.

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And then we're going to add that to, plus the derivative

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of the natural log of x.

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That's fairly straightforward, that's 1/x times x to the x.

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And of course the left hand side of the equation

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was just 1/y dy/dx.

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And we can multiply both sides of this now by y, and we get

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dy/dx is equal to y times all of this crazy stuff-- x to the

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x times the natural log of x plus 1 times the natural log of

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x plus 1/x times x to the x.

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That's x to the negative 1.

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We could rewrite this as x to the minus 1, and then

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you add the exponents.

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You could write this as x to the x minus 1 power.

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And if we don't like this y here, we can just

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substitute it back.

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y was equal to this, this crazy thing right there.

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So our final answer for this seemingly-- well on one level

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looks like a very simple problem, but on another level

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when you appreciate what it's saying, it's like oh there's a

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very complicated problem-- you get the derivative of y with

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respect to x is equal to y, which is this.

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So that's x to the x to the x times all of this stuff-- times

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x to the x natural log of x plus 1 times the natural log

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of x, and then all of that plus x to the x minus 1.

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So who would have thought.

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Sometimes math is elegant.

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You take the derivative of something like this and

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you get something neat.

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For example, when you take the derivative of natural

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log of x you get 1/x.

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That's very simple and elegant, and it's nice that math

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worked out that way.

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But sometimes you do something, you take an operation on

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something that looks pretty simple and elegant, and you get

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something that's hairy and not that pleasant to look

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at, but is a pretty interesting problem.

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And there you go.

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