second order partial derivatives

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luka16

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Hello.
How can I solve the second-order partial derivative like this one?
sa.PNG.png
This just makes me confused.
I know how to determine second-order partials for example f(x,y)=2x^4-5x^3y^3-y^4, it is very easy to solve it for fxx, fyy and fxy, but can't understand what to do in the exercise which is on the pic
can anyone help?
 
luka16 said:
Hello.
How can I solve the second-order partial derivative like this one?
View attachment 30723
This just makes me confused.
I know how to determine second-order partials for example f(x,y)=2x^4-5x^3y^3-y^4, it is very easy to solve it for fxx, fyy and fxy, but can't understand what to do in the exercise which is on the pic
can anyone help?
Click to expand...

This is a problem of implicit differentiation - basically same principle as the example function you cite.

Please show step-by-step , how you would calculate fxx for f(x,y)=2x^4-5x^3y^3-y^4,
 
Subhotosh Khan said:
f(x,y)=2x^4-5x^3y^3-y^4,
Click to expand...
f(2,3) = 2*2^4 - 5*2^3*3 * 3^3 - 3^4 = -1129

I am really surprised that you are a student of CALCULAS, yet cannot evaluate a function at a given point!!
 
Last edited by a moderator:
Subhotosh Khan said:
f(2,3) = 2*2^4 - 5*2^3*3 * 3^3 - 3^4 = -1129

I am really surprised that you are a student of CALCULAS, yet cannot evaluate a function at a given point!!
Click to expand...
I am sorry, I think you misunderstood my question. I know what to do in that case, but a question from the image is one which I wished to know how to do step-by-step, but without using f(x,y)=2x^4-5x^3y^3-y^4, I just mentioned that I know how to solve probs like f(x,y)=2x^4-5x^3y^3-y^4 this one, but I don't know how to solve a problem which is one image...
according to you I just need to write 2/1 - 1/2 and that's all?
sa.PNG.png
this is what I wish to know how to solve.
sorry for misunderstood
 
luka16 said:
okay, but what is fxx(2,1) for example?
Click to expand...
[imath]f_{xx}(2,1)[/imath] is the second partial derivative of the function evaluated at the point (2,1). Start with [imath]f_x[/imath]. Give it a try and post back your attempt, even if it's wrong. Remember what I said earlier: Treat y as a constant (like number 2) and differentiate with respect to x.
 
luka16 said:
I am sorry, I think you misunderstood my question. I know what to do in that case, but a question from the image is one which I wished to know how to do step-by-step, but without using f(x,y)=2x^4-5x^3y^3-y^4, I just mentioned that I know how to solve probs like f(x,y)=2x^4-5x^3y^3-y^4 this one, but I don't know how to solve a problem which is one image...
according to you I just need to write 2/1 - 1/2 and that's all?
View attachment 30744
this is what I wish to know how to solve.
sorry for misunderstood
Click to expand...
Have you calculated fxx for f(x,y) = (x/y) - (y/x) ?

First calculate fx for f(x,y) = (x/y) - (y/x).......................(assuming 'y' is constant)

then we will continue.....
 
luka16 said:
I am sorry, I think you misunderstood my question. I know what to do in that case, but a question from the image is one which I wished to know how to do step-by-step, but without using f(x,y)=2x^4-5x^3y^3-y^4, I just mentioned that I know how to solve probs like f(x,y)=2x^4-5x^3y^3-y^4 this one, but I don't know how to solve a problem which is one image...
according to you I just need to write 2/1 - 1/2 and that's all?
View attachment 30744
this is what I wish to know how to solve.
sorry for misunderstood
Click to expand...
If I were to ask you to differentiate this function, how would you do it? [math]f(x)=\frac{x}{2}-\frac{2}{x}[/math]
 
I realized what your confusion is. You're confused about [imath]f_{xx}(x,y)[/imath] versus [imath]f_{xx}(2,1)[/imath] notation. The latter means differentiate as you would normally. Once you have your partial derivative, plug in [imath]x=2\medspace and\medspace y=1[/imath].
 
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