Welcome to Implicit Differentiation
Using Partial Derivatives.
Remember, we performed implicit differentiation
in Calculus I.
The formula below provides an alternative method
for determining dy/dx given an implicit function.
Here's how it works.
If an equation F of x, y is equal to zero
defines y implicitly as a differentiable function of x,
then we can determine dy/dx by taking the opposite
of the partial derivative of F with respect to x
divided by the partial derivative of F with respect to y
as long as Fy does not equal zero.
So let's take a moment and see
where this formula comes from.
So we're going to start with F of x, y
equals zero.
Now, if our equation is not equal to zero,
we can go ahead and move terms around
so that is equal to zero.
Now, what we're going to do is differentiate
both sides of the equation with respect to x.
And when we do this, we're going to have
to apply the chain rule to the left side
meaning we're first going to find the partial
with respect to x.
Then multiply that by dx/dx.
Plus the partial derivative with respect to y
times dy/dx.
Then of course the right side would be zero.
Well, dx/dx is going to be equal to one.
So now what we're going to do is solve this equation for dy/dx.
We'll first subtract Fx on both sides of the equation.
That's going to give us the partial with respect to y
times dy/dx
equals the opposite of the partial of F with respect to x.
And now we just divide both sides by Fy,
and we have our formula from the previous screen.
Dy/dx is equal to the opposite
of the partial derivative of F with respect to x
divided by the partial derivative of F with respect to y.
Let's go and take a look at some examples
and we'll also compare this
to the method we learned in Calc I.
We're going to determine dy/dx of this implicit equation.
So, it's already equal to zero
so we can go ahead and state that F of x, y
is equal to y to the fourth
plus x squared y
minus 12.
So dy/dx
equals the opposite of,
our numerator is going to be the partial derivative
with respect to x.
So we'll treat y as a constant that would be zero.
This would be two xy.
This would be zero.
And our denominator is going to be the partial derivative
with respect to y.
So we have four y cubed.
This would plus x squared times one,
so plus x squared.
This would be zero.
So as you can see, it's a pretty quick easy process.
We have now found dy/dx.
Let's go and compare this method
to the method we learned in Calculus I.
That's the same equation
but now we're going to use the method from Calculus I.
So we'll find the derivative of each term,
as we normally do,
and if it has a y in it
we'll have to multiply it by an extra factor of dy/dx
because we'd be applying the chain rule.
So we differentiate both sides with respect to x.
So here we have four y cubed times dy/dx.
Now, here we're going to apply the product rule.
So it'd be plus the first x squared
times zero y, that will be one, times dy/dx
plus y
times derivative of z squared, that'd be two x.
Derivative of 12 would be zero,
and the right side would also be zero,
and now we have to solve this equation for dy/dx.
So we're going to factor out dy/dx
and then move this term to the right side.
So we'd have four y cubed
plus x squared,
and then we're going to go ahead and move
two xy to the other side, so we'll have negative two xy.
And now we divide.
So as you can see,
the first method was a lot easier and a lot faster.
Let's go ahead and make sure the results are the same.
Here's what we determined using the old technique.
I'm going back to the previous screen.
It's a lot less work, dy/dx of course is the same.
Let's go and take a look at a couple more examples.
Notice that this is not equal to zero.
So that means F of x,y
would be equal to sine x
plus tangent xy
minus 5.
And now we can go ahead and apply the formula.
So we have dy/dx
equals the opposite of this fraction
where the numerator is the partial derivative
with respect to x, so we're going to have cosine x
plus, so the derivative of tangent xy
is going to be secant squared xy
times u prime,
derivative of xy with respect to x is going to be y,
and of course derivative of five would be zero.
Let me divide this by the partial derivative of F
with respect to y.
So this derivative would be zero
and this is going to be secant squared xy
times x.
And of course derivative of five is zero again.
Let's go ahead and simplify this.
First fraction would be negative cosine x
all over x secant squared xy.
Now, for the next fraction
be careful of this negative sign here.
The entire fraction is negative
so it'd have minus y secant squared xy
all over x secant squared xy.
So this part does simplify.
So our final derivative would be
negative cosine x over x secant squared xy,
so it'd just be minus y over x.
And let's go and take a look at one more.
Let's go and write out F of xy.
Alright, this is two natural log,
x squared plus y squared to the one half power
plus three xy,
this would be minus four.
One more thing we can do is we can take this exponent here
and apply the power property of logarithms
to simplify this function.
Let's do that.
So we'd have natural log of x squared plus y squared
plus three xy minus four.
Okay, let's see if we can determine dy/dx.
Don't forget the negative in the formula.
So we're going to find the partial derivative of F
with respect to x.
Remember, the derivative of natural log u
is one over u times u prime.
So we'll have one over x squared plus y squared.
Then u prime.
The derivative of x squared plus y squared
with respect to x would be two x treating y as a constant.
Plus, the derivative of three xy with respect to x
would be just three y.
So now we're going to find the partial derivative
with respect to y.
So we'll have one over u,
or one over x squared plus y squared,
times, derivative of x squared plus y squared
with respect to y would be two y.
Plus, derivative of three xy with respect to y
would be three x.
So this is dy/dx, but you can see it's kind of messy.
So what we'll do now is clear the denominators
by multiplying both the top and the bottom of this fraction
by x squared plus y squared over one.
So let's see what this does.
We multiply this fraction by this fraction.
Notice the numerator/denominator simplify out
so we're just left with two x
plus, here we have to multiply it three y
times x squared plus y squared.
That's going to be three x squared y.
Plus three y cubed.
Let's look at the denominator now.
Again, when I multiply this fraction by this fraction here,
the numerator/denominator simplify it out
so we're left with two y.
Now if we multiply three x by x squared plus y squared,
so we'll have plus three x cubed
plus three x y squared.
Now, it's very tempting to try
to simplify the twos and the threes
and maybe even some of the variables.
But remember, we can't simplify
across addition or subtraction,
and since the numerator/denominator do not factor,
this does not simplify and we're done.
In the next video we'll take a look
at expanding this formula here
to help us determine partial derivatives
where instead of F of xy we have F of xyz.
I hope you found these examples helpful.
Thank you for watching.