Uploaded by TheIntegralCALC on 11.05.2011

Transcript:

Hi, everyone. Welcome back to integralcalc.com. Today we’re going to be doing another implicit

differentiation problem. And the equation that we’re going to be implicitly deriving

is x sin y + y sin x = 1. And we’re going to be using the product rule formula to do

it. In order to perform implicit differentiation,

remember that we have to differentiate both sides of the equation. We’ll start with

the left-hand side. We’re going to have to use product rule for both of these terms

here, x sin y and y sin x. let’s start with the first one. Our product rule formula tells

us that when we have both functions multiplied together, f(x) and g(x), taking the derivative

of those two functions multiplied together means that the result is going to look like

this: f’(x)g(x) + f(x)g’(x). Let’s assign f(x) and g(x) to parts of our term here. So

we’ll call f(x) = x and we’ll call g(x) = sin y. Basically, we just took x here as

f(x) and sin y as g(x) to match this format here of our product rule formula. It doesn’t

matter which one you pick. Either way will work with the formula. So then the formula

tells us that the first thing we need to do is write the derivative of f(x). The derivative

of f(x) is just going to be 1 because the derivative of x is 1. Remember that with implicit

differentiation, we’re taking the derivative of both x and y with respect to x. Whenever

we’ve got something where only x is involved, it’s a no-brainer. It’s just like a regular

derivative so taking the derivative of x here will just be 1.

Then we multiply by g(x) according to our formula, which we’ve written out as sin

y. Then you can see that according to the formula we’re going to add f(x) which is

just x and then we multiply by g’(x). Here’s where things get a little bit trickier. So

g(x) is sin y and we need to take the derivative of that. In order to take the derivative of

sin y, we’re actually going to need to use chain rule. We’ve said that chain rule tells

us to take the derivative of the outside function, leaving the inside function completely alone

and then after the fact, multiply by the derivative of the inside. Well, the outside function

is our sine function, sin y. The inside function is just y, what’s inside of the sine function.

So the derivative of sine is cosine. What we do is take the derivative of the outside,

cosine and leaving the inside completely alone. The inside is y. We don’t touch that at

all right now. We’re just looking at the outside and the derivative of sine is cosine.

That’s the derivative of the outside function, cos y because we leave that inside function

alone without doing anything to it yet. Then we have to multiply by the derivative

of the inside function. The inside function is y. Taking the derivative of that gives

us 1 because we’re treating y as a variable just like x, except for the fact that whenever

we take the derivative of y, we have to multiply by dy/dx because y is a function of x. So

normally, when we take the derivative of x, it’s just 1. If we take the derivative of

y with implicit differentiation, it’s also 1. We treat it as a normal variable just like

x. So we’ve got 1 here but we just took the derivative of y, we have to then multiply

by dy dx. That’s the only thing that’s special about implicit differentiation. So

you multiply by dy dx and that is the end of the product rule formula for this first

term here x sin y. Now, let’s go ahead and take the derivative

of our second term y sin x. This was the derivative of our first term: (1)(sin y) + (x)(cos y)(1)

dy/dx. Then we’re going to add to that because we’ve got addition here on our original

equation. Now, we’re taking the derivative of y sin x. We’ll say that f(x) from our

product rule formula again is equal to y and that g(x) = sin x. So remember again, first

we’ll need the derivative of f(x). The derivative of y is simply 1 because we treat it like

a normal variable just like in x but since we just took the derivative of y, we have

to multiply immediately by dy/dx. That’s the rule with implicit differentiation. So

there’s that term. Then we multiply by g(x) which is sin x then we add and we write down

f(x) which is y and then we multiply by g’(x). That’s going to be the derivative

of our sin x function. Remember we need to use chain rule again. Taking the derivative

of the outside function sine and the derivative is cosine. So we say cos x because we leave

that inside function completely alone. We’re just looking at the outside which is sine.

But then we have to multiply by the derivative of the inside function. The inside function

is x and the derivative of x is 1 so we multiply by 1. We don’t have to multiply by dy/dx

because we only do that when we’re taking the derivative of y, not the derivative of

x. So we multiply by 1 and that’s the end of the product rule formula for this term

here, y sin x. That’s the left side. And the right side

we just have 1 here. The derivative of any constant is zero so we’re going to end up

with zero on the right-hand side. So this is our full derivative: [(1)(sin y) + (x)(cos

y)(1) dy/dx] + [(1) dy/dx (sin x) + (y)(cos x)(1)] = 0. Now it’s just a matter of simplification.

So to simplify, it looks like we’ll get sin y + x cos y dy/dx + sin x dy/dx + y cos

x = 0. So we’ve simplified about as much as we can.

Now our next step is to get dy/dx all by itself on the left-hand side of our equation. The

way we do that is two steps first. We separate our terms so that every term that does not

include dy/dx, it’s moved to the right-hand side. And all terms that do include dy/dx

on the left-hand side. Then once we’ve got them there, we’re going to factor out dy/dx

from all those terms and then divide out that remainder so that we’re just left with dy/dx.

As you can see, sin y and y cos x is going to move to the right-hand side. On the left-hand

side, we’re just going to be left with x cos y dy/dx + sin x dy/dx. And then on the

right-hand side, we will subtract sin y from both sides so we’ll get a negative sin y

and we’ll subtract y cos x from both sides so we’ll get minus y cos x. Then, like I

said we want to factor out dy/dx on the left-hand side. Doing that leaves us with x cos y +

sin x = -sin y – y cos x. And now that we’ve factored out that dy/dx, we could take that

whole remaining chunk, (x cos y + sin x) and move that to the right-hand side by dividing

both sides by that quantity. What that’s going to do is give us our final answer. As

soon as you get dy/dx all by itself on the left-hand side and you’ve simplified by

the right-hand side as much as possible, that’s your final answer. So you’ll get dy/dx = (-sin

y – y cos x)/(x cos y + sin x). That’s it. That’s our final answer. I

hope that video helped you guys and I will see you in the next one. Bye!

differentiation problem. And the equation that we’re going to be implicitly deriving

is x sin y + y sin x = 1. And we’re going to be using the product rule formula to do

it. In order to perform implicit differentiation,

remember that we have to differentiate both sides of the equation. We’ll start with

the left-hand side. We’re going to have to use product rule for both of these terms

here, x sin y and y sin x. let’s start with the first one. Our product rule formula tells

us that when we have both functions multiplied together, f(x) and g(x), taking the derivative

of those two functions multiplied together means that the result is going to look like

this: f’(x)g(x) + f(x)g’(x). Let’s assign f(x) and g(x) to parts of our term here. So

we’ll call f(x) = x and we’ll call g(x) = sin y. Basically, we just took x here as

f(x) and sin y as g(x) to match this format here of our product rule formula. It doesn’t

matter which one you pick. Either way will work with the formula. So then the formula

tells us that the first thing we need to do is write the derivative of f(x). The derivative

of f(x) is just going to be 1 because the derivative of x is 1. Remember that with implicit

differentiation, we’re taking the derivative of both x and y with respect to x. Whenever

we’ve got something where only x is involved, it’s a no-brainer. It’s just like a regular

derivative so taking the derivative of x here will just be 1.

Then we multiply by g(x) according to our formula, which we’ve written out as sin

y. Then you can see that according to the formula we’re going to add f(x) which is

just x and then we multiply by g’(x). Here’s where things get a little bit trickier. So

g(x) is sin y and we need to take the derivative of that. In order to take the derivative of

sin y, we’re actually going to need to use chain rule. We’ve said that chain rule tells

us to take the derivative of the outside function, leaving the inside function completely alone

and then after the fact, multiply by the derivative of the inside. Well, the outside function

is our sine function, sin y. The inside function is just y, what’s inside of the sine function.

So the derivative of sine is cosine. What we do is take the derivative of the outside,

cosine and leaving the inside completely alone. The inside is y. We don’t touch that at

all right now. We’re just looking at the outside and the derivative of sine is cosine.

That’s the derivative of the outside function, cos y because we leave that inside function

alone without doing anything to it yet. Then we have to multiply by the derivative

of the inside function. The inside function is y. Taking the derivative of that gives

us 1 because we’re treating y as a variable just like x, except for the fact that whenever

we take the derivative of y, we have to multiply by dy/dx because y is a function of x. So

normally, when we take the derivative of x, it’s just 1. If we take the derivative of

y with implicit differentiation, it’s also 1. We treat it as a normal variable just like

x. So we’ve got 1 here but we just took the derivative of y, we have to then multiply

by dy dx. That’s the only thing that’s special about implicit differentiation. So

you multiply by dy dx and that is the end of the product rule formula for this first

term here x sin y. Now, let’s go ahead and take the derivative

of our second term y sin x. This was the derivative of our first term: (1)(sin y) + (x)(cos y)(1)

dy/dx. Then we’re going to add to that because we’ve got addition here on our original

equation. Now, we’re taking the derivative of y sin x. We’ll say that f(x) from our

product rule formula again is equal to y and that g(x) = sin x. So remember again, first

we’ll need the derivative of f(x). The derivative of y is simply 1 because we treat it like

a normal variable just like in x but since we just took the derivative of y, we have

to multiply immediately by dy/dx. That’s the rule with implicit differentiation. So

there’s that term. Then we multiply by g(x) which is sin x then we add and we write down

f(x) which is y and then we multiply by g’(x). That’s going to be the derivative

of our sin x function. Remember we need to use chain rule again. Taking the derivative

of the outside function sine and the derivative is cosine. So we say cos x because we leave

that inside function completely alone. We’re just looking at the outside which is sine.

But then we have to multiply by the derivative of the inside function. The inside function

is x and the derivative of x is 1 so we multiply by 1. We don’t have to multiply by dy/dx

because we only do that when we’re taking the derivative of y, not the derivative of

x. So we multiply by 1 and that’s the end of the product rule formula for this term

here, y sin x. That’s the left side. And the right side

we just have 1 here. The derivative of any constant is zero so we’re going to end up

with zero on the right-hand side. So this is our full derivative: [(1)(sin y) + (x)(cos

y)(1) dy/dx] + [(1) dy/dx (sin x) + (y)(cos x)(1)] = 0. Now it’s just a matter of simplification.

So to simplify, it looks like we’ll get sin y + x cos y dy/dx + sin x dy/dx + y cos

x = 0. So we’ve simplified about as much as we can.

Now our next step is to get dy/dx all by itself on the left-hand side of our equation. The

way we do that is two steps first. We separate our terms so that every term that does not

include dy/dx, it’s moved to the right-hand side. And all terms that do include dy/dx

on the left-hand side. Then once we’ve got them there, we’re going to factor out dy/dx

from all those terms and then divide out that remainder so that we’re just left with dy/dx.

As you can see, sin y and y cos x is going to move to the right-hand side. On the left-hand

side, we’re just going to be left with x cos y dy/dx + sin x dy/dx. And then on the

right-hand side, we will subtract sin y from both sides so we’ll get a negative sin y

and we’ll subtract y cos x from both sides so we’ll get minus y cos x. Then, like I

said we want to factor out dy/dx on the left-hand side. Doing that leaves us with x cos y +

sin x = -sin y – y cos x. And now that we’ve factored out that dy/dx, we could take that

whole remaining chunk, (x cos y + sin x) and move that to the right-hand side by dividing

both sides by that quantity. What that’s going to do is give us our final answer. As

soon as you get dy/dx all by itself on the left-hand side and you’ve simplified by

the right-hand side as much as possible, that’s your final answer. So you’ll get dy/dx = (-sin

y – y cos x)/(x cos y + sin x). That’s it. That’s our final answer. I

hope that video helped you guys and I will see you in the next one. Bye!