Implicit differentiation
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Implicit differentiation
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Showing explicit and implicit differentiation give same result
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Implicit derivative of (x-y)^2 = x + y + 1
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Implicit derivative of y = cos(5x - 3y)
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Implicit derivative of (x^2+y^2)^3 = 5x^2y^2
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Finding slope of tangent line with implicit differentiation
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Implicit derivative of e^(xy^2) = x - y
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Derivative of x^(x^x)
Derivative of x^(x^x) Derivative of x^(x^x)
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- A bit of a classic implicit differentiation problem
- is the problem y is equal to x to the x.
- And then to find out what the derivative of y
- is with respect to x.
- And people look at that, oh you know, I don't have just a
- constant exponent here, so I can't just use the power
- rules, how do you do it.
- And the trick here is really just to take the natural log of
- both sides of this equation.
- And this is going to build up to what we're going to
- do later in this video.
- So If you take the natural log on both sides of this equation,
- you get the natural log of y is equal to the natural
- log of x to the x.
- Now our power rules, or I guess our natural log rules, say
- look, if I'm taking the natural log of something to the
- something, this is equivalent to, I can rewrite the natural
- log of x to the x as being equal to x times the
- natural log of x.
- So let me rewrite everything again.
- If I take the natural log of both sides of that equation, I
- get the natural log of y is equal to x times the
- natural log of x.
- And now we can take the derivative of both sides of
- this with respect to x.
- So the derivative with respect to x of that, and then
- the derivative with respect to x of that.
- Now we're going to apply a little bit of the chain rule.
- So the chain rule.
- What's the derivative of this with respect to x?
- What's the derivative of our inner expression
- with respect to x?
- It's a little implicit differentiation, so it's dy
- with respect to x times the derivative of this whole
- thing with respect to this inner function.
- So the derivative of natural log of x is 1/x.
- So the derivative of natural log of y with
- respect to y is 1/y.
- So times 1/y.
- And the derivative of this-- this is just the product rule,
- and I'll arbitrarily switch colors here-- is the derivative
- of the first term, which is 1, times the second term, so times
- the natural log of x plus the derivative of the second term,
- which is 1/x times the first term.
- So times x.
- And so we get dy/dx times 1/y is equal to natural log of x
- plus-- this just turns out to be 1-- x divided by x, and
- then you multiply both sides of this by y.
- You get dy/dx is equal to y times the natural
- log of x plus 1.
- And if you don't like this y sitting here, you could just
- make the substitution.
- y is equal to x to the x.
- So you could say that the derivative of y with respect to
- x is equal to x to the x times the natural log of x plus 1.
- And that's a fun problem, and this is often kind of given as
- a trick problem, or sometimes even a bonus problem if people
- don't know to take the natural log of both sides of that.
- But I was given an even more difficult problem, and
- that's what we're going to tackle in this.
- But it's good to see this problem done first because it
- gives us the basic tools.
- So the more difficult problem we're going to
- deal with is this one.
- Let me write it down.
- So the problem is y is equal to x to the-- and here's the
- twist-- x to the x to the x.
- And we want to find out dy/dx.
- We want to find out the derivative of y
- with respect to x.
- So to solve this problem we essentially use the same tools.
- We use the natural log to essentially breakdown this
- exponent and get it into terms that we can deal with.
- So we can use the product rule.
- So let's take the natural log of both sides of this equation
- like we did last time.
- You get the natural log of y is equal to the natural log
- of x to the x to the x.
- And this is just the exponent on this.
- So we can rewrite this as x to the x times the natural log
- times the natural log of x.
- So now our expression our equation is simplified to the
- natural log of y is equal to x to the x times the
- natural log of x.
- But we still have this nasty x to the x here.
- We know no easy way to take the derivative there, although I've
- actually just shown you what the derivative of this is, so
- we could actually just apply it right now.
- I was going to take the natural log again and it would turn
- into this big, messy, confusing thing but I realized that
- earlier in this video I just solved for what the
- derivative of x to the x is.
- It's this thing right here.
- It's this crazy expression right here.
- So we just have to remember that and then apply and
- then do our problem.
- So let's do our problem.
- And if we hadn't solved this ahead of time, it was kind of
- an unexpected benefit of doing the simpler version of the
- problem, you could just keep taking the natural log of this,
- but it'll just get a little bit messier.
- But since we already know what the derivative of x to the
- x is, let's just apply it.
- So we're going to take the derivative of both
- sides of the equation.
- Derivative of this is equal to the derivative of this.
- We'll ignore this for now.
- Derivative of this with respect to x is the derivative of
- the natural log of y with respect to y.
- So that's 1/y times the derivative of y
- with respect to x.
- That's just the chain rule.
- We learned that in implicit differentiation.
- And so this is equal to the derivative of the first term
- times the second term, and I'm going to write it out here just
- because I don't want to skip steps and confuse people.
- So this is equal to the derivative with respect to x of
- x to the x times the natural log of x plus the derivative
- with respect to x of the natural log of
- x times x to the x.
- So let's focus on the right hand side of this equation.
- What is the derivative of x to the x with respect to x?
- Well we just solved that problem right here.
- It's x to the x natural log of x plus 1.
- So this piece right there-- I already forgot what it
- was-- it was x to the x natural log of x plus 1.
- That is x to the x times the natural log of x plus 1.
- And then we're going to multiply that times
- the natural log of x.
- And then we're going to add that to, plus the derivative
- of the natural log of x.
- That's fairly straightforward, that's 1/x times x to the x.
- And of course the left hand side of the equation
- was just 1/y dy/dx.
- And we can multiply both sides of this now by y, and we get
- dy/dx is equal to y times all of this crazy stuff-- x to the
- x times the natural log of x plus 1 times the natural log of
- x plus 1/x times x to the x.
- That's x to the negative 1.
- We could rewrite this as x to the minus 1, and then
- you add the exponents.
- You could write this as x to the x minus 1 power.
- And if we don't like this y here, we can just
- substitute it back.
- y was equal to this, this crazy thing right there.
- So our final answer for this seemingly-- well on one level
- looks like a very simple problem, but on another level
- when you appreciate what it's saying, it's like oh there's a
- very complicated problem-- you get the derivative of y with
- respect to x is equal to y, which is this.
- So that's x to the x to the x times all of this stuff-- times
- x to the x natural log of x plus 1 times the natural log
- of x, and then all of that plus x to the x minus 1.
- So who would have thought.
- Sometimes math is elegant.
- You take the derivative of something like this and
- you get something neat.
- For example, when you take the derivative of natural
- log of x you get 1/x.
- That's very simple and elegant, and it's nice that math
- worked out that way.
- But sometimes you do something, you take an operation on
- something that looks pretty simple and elegant, and you get
- something that's hairy and not that pleasant to look
- at, but is a pretty interesting problem.
- And there you go.
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