- #1

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Z=y+x^2*y+x^2+x^3+x^4+5

I would like to find the partial derivative of:

diff(z,x) ?

diff(z,y)?

Kindly give me a step by step solution.

Hope to hear from you soon. Thanking you all in advance for your replies.

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- Thread starter rovaniemi
- Start date

- #1

- 10

- 0

Z=y+x^2*y+x^2+x^3+x^4+5

I would like to find the partial derivative of:

diff(z,x) ?

diff(z,y)?

Kindly give me a step by step solution.

Hope to hear from you soon. Thanking you all in advance for your replies.

- #2

Mentallic

Homework Helper

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What does it mean to take the partial derivative of a function?

- #3

UltrafastPED

Science Advisor

Gold Member

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Thus for diff(Z,x) you simply treat y as a constant and then differentiate wrt x as normal:

diff(Z,x) = 2xy + 2x + 3x^2 +4x^3.

Now you do diff(Z,y)!

- #4

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Thus for diff(Z,x) you simply treat y as a constant and then differentiate wrt x as normal:

diff(Z,x) = 2xy + 2x + 3x^2 +4x^3.

Now you do diff(Z,y)!

- #5

CAF123

Gold Member

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Orthogonal coordinates are a special case of curvilinear coordinates. A set of orthogonal coordinates means that the coordinates that make up the basis of your coordinate system are, well, orthogonal. The most obvious example (and the case in question) is the Cartesian system defined by the ##\left\{i,j,k\right\}## basis.Could please explain what you meant by orthogonal corodinates?

Other examples you may have seen include spherical and cylindrical coordinates.

- #6

Mentallic

Homework Helper

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I would be highly grateful if you can show me step by step. Thank you.

It's easier than it seems!

If you have the function

y = 3 + x^2*3 + x^2 + x^3 + x^4 + 5

Then what is dy/dx?

Now, I'm sure you can find that, so once you do, since you considered 3 to be a constant, do the same for y when taking the partial derivative dz/dx of the function

Z=y+x^2*y+x^2+x^3+x^4+5

3 is just like y. You treat it like a constant. Then to find dz/dy, just treat x as a constant.

- #7

chiro

Science Advisor

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If two vectors are orthogonal then they are independent and this translates to the situation where changing one vector won't have any impact on changing the other.

In a non-orthogonal situation, changing one quantity will change another in the general situation.

- #8

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It's easier than it seems!

If you have the function

y = 3 + x^2*3 + x^2 + x^3 + x^4 + 5

Then what is dy/dx?

Now, I'm sure you can find that, so once you do, since you considered 3 to be a constant, do the same for y when taking the partial derivative dz/dx of the function

Z=y+x^2*y+x^2+x^3+x^4+5

3 is just like y. You treat it like a constant. Then to find dz/dy, just treat x as a constant.

Thank you so much for the reply.

When

y = 3 + x^2*3 + x^2 + x^3 + x^4 + 5

Should the answer be like this:

dy/dx =6x +2x+3x^2+4x^3 when i consider 3 as constant and is it like 5(number alone without x or y variable) should also be ignored.

- #9

HallsofIvy

Science Advisor

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What they mean is that z= 3xy+ 2x^2- 4xy^2+ x^2y^2, would be thought of as

1) (differentiating with respect to x) z= Ax+ Bx^2- Cx+ Dx^2, where A= 3y, B= 2, C= 4y^2, and D= y^2 are "constants", so that z_x= A+ 2Bx- C+ 2Dx= 3y+ 4x- y^2+ 2y^2x.

2) (differentiating with respect to y) z= Ay+ B- Cy^2+ Dy^2, where now A= 3x, B= 2x^2, C= 4x, and D= x^2 are "constants", so that z_y= A- 2Cy+ 2Dy= 3x- 2x^2y+ 2x^2y.

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