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Problem 16

Suppose $ f $ is a differentiable function of $ x…

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Problem 15

Suppose $ f $ is a differentiable function of $ x $ and $ y $, and $ g(u, v) = f(e^u + \sin v, e^u + \cos v) $. Use the table of values to calculate $ g_u(0, 0) $ and $ g_v(0, 0) $.

Answer

$$7,2$$



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Video Transcript

Hi there. In this problem, we have function g of the variables you under the and we can also think of it as a function f the variables X and X depends on you and V in sodas. A. Why keep this in mind as we go through here? Because the variables could be tricky. So if we make a tree diagram, it's in the book with F depends on variables X and Y X and Y each depend on the variables you and be, uh, so let's use this to get our answer as we go. So first we were asked to find g partial with respect to you the 0.0 comma zero. Now that means when you and V. R. Zero since you envy, are the inputs for Gene. So to get this let's use the tree diagram here, get down to the variable you. So if you go down the left branch, get F partial X and then we multiply by experts, will you plus good on the other side of partial. Why times why partial you and you want to evaluate this whole thing when you and V are both zero. Okay, so let's go carefully left, right and value each of these f of x m not f of x fxfx here f partial with respect to X. When you and the are both zero, we need to figure out what perhaps inputs are. In other words, uh, when u and V are both zero, what is X? So it's look and X, our first input for F it defined as each of you plus Sign V. And so let's keep track of this. Pecs equals eat of the U plus Sign V and why the second and put either the U Plus Co sign Be so when you enviable zero. Let's write what X and wire so we don't get confused. X equals well. We get eat of zero plus sign. Zero. You know zero's one sign of zero is 01 plus zero is one. Why equals? Let's see either zero plus co. Sign. Zero. You know, zero's one co sign. Zero is also 11 of us wants to X and wire to this makes sense. Looking at the table they gave us pretty much had to be that, but it's always good to check. So now we know exactly where to look in the table for each of these. You want f sub x, and this is at the point where X and wire wanted to. So if you look in the table, we see that f partial X at one comma to as the bottom row would give us the number two. And so that gives us two now x sub you comes next and an ex eyes given in the U. N V variables. So this will get a top row of the table. Um, and X sub you. I'm sorry, we actually don't need the table for this one. It's staring us right in the face x sub you. Let's calculate that right here. So x sub you. Here's our exit over there. So the partial derivative eat of the U plus sign of V is neither the U. S the derivative eat of the U and sign of he has no using it. So that's it, Um, and in at zero comma zero, you was really just zero. So that will just be the number one. Okay, next up, f partial of why Same thing is before it's looking a table. We care about the point one comma two That will be the bottom room. And we see that's five. So and from the table, partial lies five. And why partial you want? Same thing is over here. Let's look why partial you is again. Eat of the U And when you and beer zero that becomes this. So they get two times one plus five times one two plus five equals seven. So that's our first answer. Okay. The exact same idea now for, uh, G partial the at zero common zero. Um, this time we're just going to go down the tree diagram again. So you want to get down to the variable V this time, though, So we get f partial X times X partial V plus of personal. Why times Why apart for V? Once again, we are evaluating this zero comma zero or you envy on. That means that our point x y again, it's one to have none of that changes. Okay, so let's use a combination of the table and the partial derivatives here. Um, the first I think the 1st 2 that we got don't change the red numbers up in our first. Our first problem here. Don't change f partial acts. At this point, it's still too and for that matter of powerful wise still five. None of that changes what does change you want? X partial V So let's look over here. The definition of excess heat of you plus sign V. So only the sign V will you contribute to the A partial V derivative derivative of Sign V is co sign V Coast on V. But at this point of the equal zero finally, why partial V? Let's look here. Same deal. Usually you post coast on V. The derivative of that with respect to V, is just a minus. Sign V. Any envy is zero. So we get, Let's see, two times coastlines zero is one. So this is too five times we'll sign of zero is zero. So that will just be 02 is our final answer here. Hopefully, this helps

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