How to Do Implicit Differentiation (NancyPi) - YouTube

Hi guys, I'm Nancy

and I'm going to show you how to do implicit differentiation.

What is that? Well...

up until now, you've probably been doing

explicit differentiation, which is just the normal kind of differentiation.

Explicit differentiation is when you just have a normal "y =" function.

y = some x expression

so it's a function explicitly in terms of x

and you probably already know how to take the derivative of something like this

you would just use the Power Rule...

differentiate with respect to x...

and your derivative, or dy/dx...

would just be 2x.

But what if your function is not written as "y =" some expression?

Well then you have to use something called implicit differentiation.

So if your function is not written as "y ="

but is instead written implicitly as a function of x, something like...

x squared plus y squared equals another term or a number, 16...

you can't do the normal explicit differentiation, you have to use

implicit differentiation

which is basically just a special kind of the Chain Rule.

So let me show you the steps, the main steps for implicit differentiation.

OK, so these are the two main steps for implicit differentiation

It's really always just these two.

The first step is to differentiate both sides with respect to x, so...

take the derivative in terms of x, but...

the difference is that if you differentiate a term that has y in it

you will need to attach a dy/dx to the term, multiply the term by dy/dx.

For instance, say that you wanted to differentiate, with respect to x,

a term, a y term...

like 2 y squared.

First, you can just use the Power Rule to differentiate normally

the 2 y squared would give you a derivative of 4y...

But because this term has a y in it, you will need to tack on, or multiply

a dy/dx at the end

Why do you do this?

Well, the short answer is that it's a form of the Chain Rule

because there is an outside function

the 2 y squared that you use the Power Rule for, and...

the way I've always thought of this is that technically there's an inside function

this y contains some x expression

the y is made up of, or composed

of x's or some x expression, so you will need to multiply by

an inside derivative to take care of that inside expression

and the inside derivative is dy/dx

so you can think of it that way.

The second step is then to just solve for dy/dx

that means get dy/dx alone on one side so that your expression

looks like dy/dx equals something.

Now there are some tricks

algebra tricks that you may need

to do this, and I'll show you those so you can manipulate it

to always get dy/dx equals something, as your solution.

So let me show you how it works

OK, what if you have a function like this

that doesn't start with "y ="

it's written implicitly in terms of x...

so it's a great candidate for implicit differentiation

in fact, you have to do it that way

The first thing is, just differentiate both sides with respect to x

So, for the first term, x squared

The derivative of that would just be 2x, using the Power Rule...

move on to the second term, y squared

remember I said to be on the lookout for terms with y

well, in this case, since there's a y in this term, you will need to attach a dy/dx

to the derivative, so first...

take the derivative using the Power Rule. The derivative of y squared is 2y

that's the derivative of the outside function

but because there's a y in this term

and y contains some x expression, potentially...

you will need to attach, or multiply, this term by dy/dx.

This is the difference between implicit and explicit differentiation, the normal differentiation.

Moving on, you do need to differentiate both sides of the equation, which may not be something you're used to

but if we look at the right-hand side, since we just have a constant, a number, 36

the derivative of that would be 0, as it always is for a constant

so we just write 0.

Alright, that was the first step. Differentiate both sides in terms of x.

Remember, the second step is to then for dy/dx. Get it alone.

So if there are ever any parentheses, you'd want to distribute or multiply out to get rid of them

There are none in this case

We just, we need to get the term that has dy/dx alone on one side

you need to subtract out any terms that don't have dy/dx

so we'll need to subtract the 2x...

from both sides, and we have 2 y dy/dx

equals negative 2x

In order to get dy/dx alone on the left-hand side

you'll need to divide out anything that you don't want there

namely, we don't want 2y

so we'll divide both sides by 2y. Those cancel...

and we are left with just dy/dx on the left-hand side, which is what we wanted

equals negative 2x over 2y, which can be simplified by dividing the 2's

as negative x over y

which can't be simplified any further. We have only dy/dx alone on the left-hand side

so we're done. This is your answer

for dy/dx.

OK, what is you have a function like this one?

It's a little more complicated. You'll need to use the Product Rule and the Chain Rule

but it starts the same way

Take the derivative of both sides with respect to x

The derivative of this first term, sin x, is just cos x, as you probably know

There are no y's to worry about in that term, so we just write cos x

we're going to subtract some derivative of this term

Alright, this term, x squared y, has a y in it

so that should set off your y detector.

You'll need to multiply by dy/dx at some point in this term.

Since it is a product, x squared times y

you will need to use the Product Rule. People often forget this part

and the easiest thing to do

is to use parentheses around what you get with the Product Rule.

This will make your life easier, and you will get all the right signs for your terms.

So subtract, open parentheses, using the Product Rule for x squared y

we have the first function, first function, x squared

times the derivative of the second, derivative of y is 1

But because this is implicit differentiation, we multiply by dy/dx, so it's

1 dy/dx, which is dy/dx

using the Product Rule we add the derivative of the first function. The derivative of x squared is just 2x

times the second function, y.

We're not taking the derivative of y, so we don't need to tack on a dy/dx.

It was just the function itself, y.

Close the parentheses, you're done with the Product Rule

move on to the next term, so we're adding

the derivative of y is dy/dx

and that equals the derivative of the right-hand side, derivative of 10x

with respect to x, is just 10

OK, and you're done with the first step.

Alright, we move on to the second part of this problem

which is sometimes more involved. Finding dy/dx

and getting it alone on one side, so...

if there are ever parentheses in your expression, first thing is to get rid of them.

That means distribute the negative sign, expand inside the parentheses

so you have cosine of x. You're going to distribute the negative so you have minus x squared dy/dx.

The minus applies to the second term as well because of the parentheses. That's very important. People forget that.

minus 2xy, and we still have these other terms, dy/dx plus dy/dx equals 10

Alright, no parentheses left

The next thing you want to do is get it so that you only have dy/dx terms on one side

this term, you can keep on the left side. This term, you can keep on the left side

but cosine x and negative 2xy

don't have a dy/dx in them, so you want to get rid of them

move them to the right-hand side

that means subtracting them off or adding them so that they move to the other side

in this case, we subtract cos x and we add 2xy to both sides

and that leaves us with...

Great, you only have dy/dx terms on one side

In order to get dy/dx alone, it needs to appear one time on that side

there's a trick to make this happen. It's not immediately obvious

but the trick is to factor out the dy/dx from both terms

I know you know how to factor

see, the value in that was that now

you only have dy/dx appearing one time in this whole equation

and it will make it very easy to get it alone, because

this negative x squared plus 1 that you don't want on the left-hand side, you can just divide out

like you did in the last one

and your final answer is dy/dx, alone on one side

equals 10 minus cos x plus 2xy, all over negative x squared plus 1.

That looks like a mess, but that is exactly correct. You have x's, you have y's

so this is your answer for the implicit differentiation, dy/dx.

Great, so now let me just warn you

about a few things that tend to trip people up

sometimes so that you can watch out for them.

One is that people forget to use the Product Rule sometimes

this looks really innocent, you might think that you could just write 2xy

or treat y like a constant.

No, you have to use the Product Rule, like here....

and yes, it makes it messier, but you will get the right answer.

Another thing that trips people up, they forget to factor

if the dy/dx appears more than once, toward the end.

And then one other thing that might confuse you is sometimes dy/dx

is instead written as y prime. It's the same exact thing, just different notation.

I hope this video helped you figure out implicit differentiation

'cause I just did a bunch of calculus

and it's not my job.

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