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# Implicit differentiation (advanced example)

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.