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## Differential Calculus

### Course: Differential Calculus > Unit 3

Lesson 4: Implicit differentiation (advanced examples)# Implicit differentiation (advanced example)

Implicit differentiation of y = cos(5x - 3y). Created by Sal Khan.

## Want to join the conversation?

- In this video at2:40as well as in the last video, sal subtracts something from both sides that contains dy/dx, in order to get all dy/dx to the same side. But I don't get how one of the dy/dx becomes 1. In this video it results in (1-3sin(5x-3y))dy/dx. How does dy/dx-3sin(5x-3y)*dy/dx= (1-3sin(5x-3y))*dy/dx(21 votes)
- He factored the dy/dx out. For example, in the expression x+5x we can factor out the x to get (1+5)x. The 1 is left behind to show that we had an x there before we factored. Another way to see this is to try multiplying the dy/dx back into the expression. If there was no 1 sitting there than we would have magically lost the dy/dx term in the original expression.(42 votes)

- Is this still counted as implicit derivatives? I'm confused because I thought explicit derivatives was where y was isolated on one side. What's the difference between the two?(11 votes)
- The difference is that we have y terms on
**both**sides of the equation (as y is part of the argument of the cos function). Although we have y on its own on the left-hand side, this is**not**the equation for y as a function of x. Thus, implicit differentiation is called for.(12 votes)

- Can I treat dy/dx as a fraction?

Can I multiply both sides by dx/dy to cleat the dy/dx?

I am not sure why one would do this, I just want to know if dy/dx is now its own thing or is it still a fraction.(2 votes)- no no no! that would be a HUGE abuse of notation! if this is how you are thinking, then start using the y' (read y prime) notation.(14 votes)

- Hello

I multiply the two paranteses before I multiply with -sin.

-sin (5x-3y)(5-3dy/dx) = -sin25x + sin 15x*dy/dx + sin15y - 9ydy/dx

My answer would then be :

dy/dx = (-sin25x+sin15y)/(-sin15x+1+9y)

Is this correct? How come Sal uses -5 sin (5x) and not -sin25x.?(3 votes)- No. The expression sin(5x-3y) means that you are taking the sine function of (5x-3y), That is not a multiplication and you certainly don't use the distributive property.

Thus, sin (kx) is NOT sine times kx. sin (kx) = 1/2 {ie^(-kxi)- ie^(kxi)}

Instead, you treat sin(5x-3y) as a single entity. You can break that up using a trigonometric identity, but that is not a more simple form.

For reference sake, though:

sin(5x-3y)=cos(3y)sin(5x)-cos(5x)sin(3y)

Finally, remember that k( sin x) is NOT equal to sin kx.(10 votes)

- At2:33he substracted 3sin(5x-3y) from 1 and at the end he got -5sin(5x-3y) / (1-3sin(5x-3y)). However if we substract the other way and get 3sin(5x-3y)-1 we can get 5sin(5x-3y) / (3sin(5x-3y)-1) . Would it not be right as well?(4 votes)
`-5sin(5x - 3y)/(1 - 3sin(5x-3y)) = 5sin(5x - 3y)/(3sin(5x - 3y) - 1)`

.

We 're just multiplying by`-1`

on top and bottom, and`a/b`

=`ka/kb`

for`k ≠ 0`

.(3 votes)

- At first, I got confused. However, I came to this conclusion:
`d/dx`

is the differential operator (similar to`+, -, *, /`

).`dy/dx = d/dx (y)`

is the variable (similar to`a = 5, b = 100, c = -0.12`

); so we can apply operators to it, right?(5 votes) - I was just curious as to why we must again solve for dy/dx, since that is already what we have once we use the chain rule(1 vote)
- The problem is that you had dy/dx on both sides of the equation, and the goal was to find the derivative of y with respect to x. You need the dy/dx isolated for the same reason you don't leave a linear equation as y=2x-y. It makes it much simpler to do any follow up work if you needed the equation if it's already prepared for you. This way when you plug in values for your variables you don't need to do any more algebra, it's just simple order of operations. I hope that clarified everything a little for you.(8 votes)

- How to know that the equation given to us will be solved by applying product rule or chain rule ?

Please explain with the examples if possible .

Thank you .(2 votes)- most of the time you can use the product rule when you see two variables multiplying eachother. For example like 3x4x^2. Usually a parentnesis is involved when you have to use the Chain Rule. Like (x+6)^9. Expanding it would take too much work, so you use the Chain Rule.(6 votes)

- would the answer -5sin(5x-3y)/4 be right as well?(2 votes)
- I don't think so. If we plug in a point that we know works for the original equation, say (pi/10, 0) we end up with two different answers for dy/dx: -5/4 for your expression and 5/2 for the expression in the video. However, I don't know if this is a kosher way to test.(3 votes)

- At around2:55: Why do you subtract 3 sin(5x-3y) from both sides?(2 votes)
- You want to get all values of dy/dx isolated on one side.(4 votes)

## Video transcript

Let's say we have
the relationship y is equal to cosine
of 5x minus 3y. And what I want to find
is the rate at which y is changing with respect to x. And we'll assume that
y is a function of x. So let's do what we've
always been doing. Let's apply the
derivative operator to both sides of this equation. On the left-hand
side, right over here, we get dy/dx is equal to-- now
here on the right hand side, we're going to apply
the chain rule. The derivative of the
cosine of something, with respect to
that something, is going to be equal to negative
sine of that something. So negative sine of 5x minus 3y. And then we have
to multiply that by the derivative of that
something with respect to x. So what's the derivative of the
something with respect to x? Well the derivative of
5x with respect to x is just equal to 5. And the derivative of
negative 3y with respect to x is just negative 3 times dy/dx. Negative 3 times the derivative
of y with respect to x. And now we just need
to solve for dy/dx. And as you can see, with some of
these implicit differentiation problems, this is the hard part. And actually, let me make
that dy/dx the same color. So that we can keep
track of it easier. So this is going to be dy/dx. And then I can close
the parentheses. So how can we do it? It's just going to be a little
bit of algebra to work through. Well, we can distribute
the sine of 5x minus 3y. So let me rewrite everything. We get dy-- whoops, I'm going
to do that in the yellow color-- we get dy/dx is equal
to-- you distribute the negative sine
of 5x minus 3y. You get-- so let me make sure
we know what we're doing. It's going to be, we're
going to distribute that, and we're going to
distribute that. So you're going to have
5 times all of this. So you're going to have
negative this 5 times the sine of 5x minus 3y. And then you're going to have
the negative times a negative, those are going to, you're
going to end up with a positive. And so you're going to
end up with plus 3 times the sine of 5x minus 3y dy/dx. Now what we can do
is subtract 3 sine of 5x minus 3y from both sides. So just to be clear, this
is essentially a 1 dy/dx. So if we subtract this from
both sides, we are left with-- So on the left-hand side,
we're going to have a 1 dy/dx, and we're going to subtract
from that 3 sine of 5x minus 3y dy/dx's. So you're going to
have 1 minus 3-- I'll keep the color
for the 3 for fun-- 3 sine of 5x minus 3y dy/dx's
on the left-hand side, is going to be
equal to, well, we subtracted this from both sides. So on the right-hand side,
this is going to go away. So we're just going to be
left with a negative 5 sine of 5x minus 3y. And we're in the
home stretch now. To solve for dy/dx,
we just have to divide both sides of the
equation by this. And we are left with dy/dx is
equal to this thing, negative 5 times the sine of 5x minus 3y. All of that over 1 minus
3 sine of 5x minus 3y. And we are done.