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Find the first partial derivatives of the function.$$f(x, y)=x^{y}$$...

Question

Find the first partial derivatives of the function.$$f(x, y)=x^{y}$$

Find the first partial derivatives of the function. $$f(x, y)=x^{y}$$



Answers

Find the first partial derivatives of the function.
$ f(x, y) = x^2y - 3y^4 $

Well were given a function unrest to find the first partial derivatives of this function. The function is F. Of X T. E equals T square E to the negative X. To find the partial derivative of F. With respect to X. We hold T constant and differentiate with respect to X. So we get negative T squared E to the negative X. Likewise, to find the partial derivative with respect to walk or T. That is, I'll hold X constant and differentiate with respect to T. This is to tee times E. To the negative X.

Got it. No. Were given a function and we're asked to find the first partial birth date of dysfunction function is F. Of X. Y equals X. Over wine defined the partial derivative with respect to X. I'll hold Why? Constant? And then differentiate with respect to X. This is just one over Y. Likewise, in the parking lot of this, F with respect to Y. All hold X constant, differentiate with respect to Y. This is the bottom Y times through the top zero minus the top extends the drill into the bottom one. All over the bottom Y squared, which is negative X over why squared? Perfect victory joe. We want joe algorithm. Perfect victory.

Were given a function in her ass to find the first partial derivatives of this function function is F of X. Y equals X over X plus y square. Now the party derivative of F with respect to X, find this, I'll hold y constant and differentiate with respect to X. Using the quotient rule. So we have the bottom X plus y squared tends to drive at the top one minus the top extends the drive at the bottom By the chain rule. This is two times x plus one. Don't worry about All over the bottom. X plus Y square had a big square. This simplifies to with X plus Y squared minus two, X squared minus two. Xy. This is X squared plus two. X Y plus Y squared minus two, X squared minus two. Xy. Uber X plus y to the fourth. Yeah, that makes actually instead of doing this, we can factor out Next plus Y from the top and this is equal to X plus y minus two. X over X plus Y To the 3rd power. Which is equal to negative X plus Y over X plus Y. That's huge promise. I will use the word every day that I'm present likewise to find the partial derivative of F. With respect to why? Yeah, I'll hold X constant and differentiate with respect to Y. Again, you can close the rule. This is the bottom X plus Y squared. Considered of the top which is a constant. So zero lines. The top X considered at the bottom by the chain rule. It's two times X plus Y. All over the bottom, X plus Y squared. Now, why would you give me head joe. Now? At this point we can in fact our next plus Y. Well we can rewrite this. In fact our next plus Y. To get negative two X over X plus Y huge means I could say you gotta stop being a deadbeat father and

We are going to find here the first partial derivatives of the function F. Of X. Y. Calls the Kocian A. X plus B, Y over C X plus dy so it's a rational functions division of polynomial of degree one in X and Y. And so we first calculate the first order partial derivative of F. We spent two X and Y. So first the partial derivative of F respect to X. At any point X. Y. Is equal to no recognize that as a function of Y. All of X only. We have a Kocian because we have the variable X both in the numerator and denominator. So we applied the Kocian rule here. So that is personal narrative respect to X. Of the numerator, X plus B. Y. Well not that this way times the denominator, C X plus dy minus numerator, A X plus B. Y. Times partial derivative with respect to X. Of the denominator. See explicit Ey over the square of the denominators. The explicit why? And so this is equal to the partial derivative with respect to X. Of X plus B. Y. Here we are assuming, I didn't say that at the start with A B, C and D are constants. So the partial derivative here is a so we get a times C X bless dy minus A. X plus B Y. Times. And these parcel narrative here is C. Yeah. And that oversee explicit ey square to get this. Okay. And uh that can be simplified the numerator and we developed the products here and here distributive law identification respect to the sum. And so we get a C. X plus a dy minus eight X minus C. B. Y. Over C. X. Plus the Y. Square. And we noticed that A. C. X cancel out. And then we have white common factor between these terms and this one. So we get 80 minus C. B. Times Y over C. X. Plus T. Y. Square. And so we can say that the partial derivative of f respect eggs at any point X. Y. Is equal to a. D minus C. B. Y. Over C. X. See eggs plus you I. Square. And that's our first partial derivative respect to X. Now we go to the partial derivative respect why at any point And that's equal to And again we see that we have a Kocian of functions depending on why. So we apply again the rule of derivative. Epic ocean. So the relative of the numerator. What? Mm. We respect to why that is servant of respect why of X plus B. Y. My time's the denominator. See eggs plus the Y minus the numerator. S. Is A. X plus B. Y. Times the person's derivative respect to Y. Of that. The nominators say explicitly why and all that over the square of the denominators. The explicit ones. Sure. Yeah. And now we calculated the relatives appear in the numerator. We get preservative here is B. So we get B. Time C. X. Just do I minus A X. Plus B Y. Times. And this partial derivative here is the and that oversee eggs less dy square. And so this is equal to and develop the numerator. We get B C X, blessed B D y -1X. Mhm 80 X minus mhm B D Y over C eggs plus dy square. Let me see here, check something. It's okay. So now this is equal to and now we can see that the terms video y minus video, I cancel out. So we get bc X. Mhm -80 x. But yeah, so, but the terms have a common factor X. Yeah. So we get it up and get in the numerator pc minus 80 X. And that over C X plus dy square. Yes. So what we get here is that the partial derivative of F. We respect to why at any point X Y is equal to B c minus 80 times X over c eggs. Blust. Ey yeah. Square. And I put that as the second result, wow. And then you can see we have a similar equation and in fact the coefficient in both derivatives, we can see here this coefficient here, they see when CB is this same coefficient but with opposite signs we take a common factor negative in any of this we get the other. So they are, there is a relationship between the efficient of the person derivative in the numerator. But the variable is that variable which we are not find the derivative respect to this case. Why is constant? Because we are calculating the partial derivative of F respect to X. And in this case X is a constant and eggs appears here in the numerator. And while I appears here in the memory now we could find if we want the partial derivative of second order, I'm going to do one of them. For example, partial derivatives of derivative of F respect to X twice. It means that we got to find a personal derivative respect to X of the first derivative of F respect to X. That is we get to find a partial derivative respect eggs Of the expression we got here above this one. A demon C B. Y over C X plus dy square. And it's very important to notice that in this case the numerator doesn't depend on X. So it's a constant. So we get that this is equal to the numerator. He did -9. Why times a partial derivative respect to X. Of one over C X plus dy square. Okay. Yeah. Yeah. And the denominator now depends, it does depend on X. So this is equal to a d minus C B. Y. Times. And we were going to write the fraction here as a negative exponents of see explosive Y. That is this expression one over the explosive was where is equivalent to see X plus D Y to the negative too. It's the same thing. So written this way we can apply the rules of derivatives instead of of the power instead of caution spirit is much more easier this way. So we get a d minus C B. Y Times. So so we have negative two times the same Basie explicitly. Y to the negative two minus one. We have applied here the derivative of a constant power of a function as a variable. And because the base see explosive Y is not only the variable but it's complex expression. We got applied the chain rule and multiply that by the conservative party conservative respect to X of C X. Let's see why. And we finally get here A D minus C. B of value. Why times negative two, C X plus D. Y to the negative three. And this partial derivative here is C. And arranging things here we get the numerator -2 times. See we have this negative to here see here then we have this factor A D minus C. B. In that times. Why? Mhm. Okay. Why? Yeah. And then uh that over this power here comes to denominator as C. X blessed. Ey cute. And that's a result that is well we found here instead partial derivative of F. We respect to X twice there is a second order partial derivative respect to eggs at any point is equal to negative to see A D minus C. B. Y over C. X. List the white cube. And like this we can do the others Partial derivatives of 2nd order. And that's a key idea for the second order partial derivative respect why. We applied the idea that the numerator doesn't depend why and so on. So this is the first derivative of the function.


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