Partial Derivatives Example 7


Uploaded by TheIntegralCALC on 27.10.2010

Transcript:
Hi everyone! Welcome back to integralcalc.com. We’re going to be doing another partial
derivatives problem today. This one is the function f of x y equals the quantity x y
divided by x squared plus y squared. And with this problem, the first thing that we should
notice is that because we have a fraction for our function here and we’re going to
be taking derivatives we’ll absolutely need to use the quotient rule in order to find
this derivative so we’ve gone ahead and laid up the quotient rule here. And given
that we’re going to need to use quotient rule, we know that we’ll need to assign
a quantity to f of x as well as a quantity to g of x because our quotient rule formula
here dictates that we do that. This part of the formula represents our original function
so it tells us right after that that f of x is our numerator and g of x is our denominator
which is why we’ve gone ahead and assigned f of x to our numerator, x y, and g of x to
our denominator, x squared plus y squared. And if you have any other questions about
quotient rule or you need further clarification, go ahead and see that section on my website
and we’ll just stick with partial derivatives here.
So, now that we know that we’re going to need to use quotient rule, we can start with
the partial derivative with respect to x and later we’ll do the partial derivative with
respect to y. So, we’re going to go through the partial derivative with respect to x piece
by piece according to the quotient rule formula and keep in mind that each time when we have
to take a derivative where, of course, we’re going to be taking it with respect to x. So,
the first piece that we need to add to our partial derivative with respect to x according
to our quotient rule formula is the derivative of f of x here. So we go over and we look
at f of x and we see that f of x equals x y. So we need the derivative of f of x which
means we have to take the derivative of x y but with respect to x since we’re taking
the partial derivative with respect to x. So, the partial derivative with respect to
x of the quantity x y is simply y because y is a constant, it acts like a constant number,
like two or three or four, which means that it’s like a coefficient on our x variable
here, it’s like if you had… if it… with a number two here and put it in front and
you we’re taking the derivative of two x the derivative would be two and of course
that two is representing our y. So, the derivative is simply y which is why we’ve gone ahead
and placed it here in our partial derivative formula. So, next, we see that we need to
add g of x so we just go ahead and take x squared plus y squared and put it down in
our partial derivative. And then, we need the original f of x function so we add x y,
or we should say, we subtract it because in the quotient rule formula you always subtract
in the middle here. And then, we’re going to multiply by the derivative of g of x which
in this case is two x because, again, we’re taking the derivative of our g of x function
here with respect to x so we’re treating x as a variable and y as a constant. So the
derivative of x squared will be two x and the derivative of y squared, since we’re
treating y as a constant, will be zero so we multiply by two x here in our partial derivative.
And then, you can see from our quotient rule formula that the denominator of our function
will just be g of x squared so we go ahead and add this g of x equals x squared plus
y squared and we square that and we put that in the denominator. So then, we’re just
going to simplify and we distribute this y through the x squared and the y squared and
then we multiply this x y times two x to simplify. And, what we get when we finish simplifying
is y cubed minus x squared y. This x squared y here comes from combining this x squared
y and the two x squared y here we get a… we end up with a negative x squared y, we
can combine those terms. So, that’s going to be our partial derivative
with respect to x and then if we go ahead and find our partial derivative with respect
to y we’re going to do the same thing, keeping in mind that when we take derivatives this
time we’ll be taking… taking them with respect to y instead of with respect to x.
So, again, the first thing we’ll need to do is put into our partial derivative the
derivative of f of x. So, we go over here we look at f of x, we see that f of x equals
x times y. So the partial derivative of this, but this time with respect to y, is simply
x because x acts as a coefficient on the y term, we treat it as a constant, this is just
like saying, you know, seven y or something like that, because that x is a constant so
the coefficient is what we pull out of that for the derivative of f of x. So then, we…
we simply put g of x into our partial derivative here, x squared plus y squared, and then we
subtract from that f of x which is just the x y and then multiply it by the derivative
of g of x, and in this case the derivative of g of x we’ll be taking it with respect
to y. So we treat x as a constant and the derivative of this x squared term here will
just be zero, the derivative of the y squared term will be two y so we multiply here by
two y, and then we put g of x squared in the denominator of our function. So again, just
going through in simplifying, we multiply this x here by the x squared term and the
y squared term and then we multiply x y by two y and then we can combine these two terms
here, x y squared and negative two x y squared, to get a negative x y squared and we end up
with this function here. So, when we write out here our final answer,
we just take these… these two answers that we got here. And our final answers for the
partial derivative of f of x or f… sorry, the partial derivative of f with respect to
x and the partial derivative of f with respect to y up here, here at the bottom.
So, I hope that helped and I will see you guys next time. Bye!