Partial Derivatives Example 5


Uploaded by TheIntegralCALC on 25.10.2010

Transcript:
Hi everyone. Welcome back to integralCALC.com. WeÕre going to be doing another partial derivatives
problem today. This one is f of x, y equals the natural log of x squared plus y squared.
So the first thing that we need to realize with this problem is that we are doing a partial
derivative problem so we need to take the partial derivative of f with respect to x
and the partial derivative of f with respect to y.
So because we have the natural log of a quantity x squared plus y squared, that is more complicated
than the natural log function in its pure state, just the natural log of x. WeÕre going
to need to use chain rule because weÕve got x squared plus y squared which is more complicated
than x.
So we have chain rule here that we need to use and then on top of that, because weÕre
going to be taking partial derivatives, weÕre going to need the derivative formula for the
natural log function. So we have the derivative of the natural log of x equals 1 over x. So
to use both of these formulas to take the partial derivatives with respect to both x
and y, the first thing that weÕre going to do is apply chain rule.
So for the partial derivative of f with respect to x using chain rule, weÕre going to first
take the derivative of the outside function which is the natural log function and then
leaving the inside function x squared plus y squared completely alone, then we multiply
that by the derivative of the inside function.
So as you can see here, taking the derivative of the outside function ln of x, knowing that
the derivative of the natural log function is 1 over x where we end up with 1 over x
squared plus y squared meaning we leave the inside of this function here intact.
Then because our chain rule formula tells us to multiply by the derivative of that inside
function x squared plus y squared, weÕre going to multiply by the derivative of the
inside function. But because weÕre taking the partial derivative with respect to x,
the derivative of that inside function x squared plus y squared, we have to take the derivative
of that with respect to x.
So in this case, we treat y as a constant and x as our variable. So if x is our variable,
we take the derivative as we normally would and the derivative of x squared is 2x. We
treat y as a constant which means that y squared is a constant and the derivative of y squared
is just zero.
So the derivative in this case of that inside function is just 2x. So then of course we
simplify and we get our final answer here for the partial derivative of f with respect
to x which is 2x over x squared plus y squared.
Now with the partial derivative of f with respect to y, we do the exact same thing.
We take the derivative of the outside function so we say itÕs going to be 1 over x but instead
of that x, we replace it with our inside function which weÕre leaving alone when taking the
derivative of the outside. So we leave x squared plus y squared here on the bottom.
Then we multiply by the derivative of that inside, x squared plus y squared. In this
case, weÕre taking the partial derivative with respect to y. So we treat y as our variable
and x as our constant. So treating y as the variable, that y squared here in our original
problem becomes 2y and the x squared here is going to be zero because we treat x as
a constant. So x squared ends up being a constant and so multiplying by the derivative of that
inside function with respect to y just gives us 2y.
So then we simplify to get our final answer. Now you can see weÕve taken the partial derivative
of f with respect to x and we have our final answer here and then the partial derivative
of f with respect to y and we have our final answer there.
So thatÕs it. I hope it helped and I will see you guys next time. Bye.
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