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So I have $z=x^2+xy+y^2$ And I want derivative of z with respect to x assuming y is constant and professor gave us $\frac{\partial z}{\partial x}=2x+y$

But how does he found it? Does he use limit like in non partial derivative expression?

Also if I have f(x)=$x^2+xy+y^2$ where $y$ is constant do I just need to just find the derivative of each term and then adding the derivatives of all those terms to find $f'(x)$:

  • derivative $x^2$: $2x$
  • derivative $xy$: $y$
  • derivative $y^2$: $0$

$2x+y+0=f'(x)$ ?

Thanks in advance!

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    $\begingroup$ Yes, what you (and the professor presumably) did in your last paragraph is correct. $\endgroup$ Dec 15, 2013 at 15:48
  • $\begingroup$ But I think of it that way: if $\sqrt{x+y+z}doesnot=\sqrt{x}+\sqrt{y}+\sqrt{z}$ then it seems intuitive to me that it's the same thing with derivatives, i don't know that's just me $\endgroup$ Dec 15, 2013 at 15:55
  • $\begingroup$ it works for derivatives. $(u+v)'=u'+v'$. That's one of the basic properties of derivation. Does not work for product though. $(uv)'=u'v+v'u\neq u'v'$ $\endgroup$
    – imj
    Dec 15, 2013 at 15:57
  • $\begingroup$ imj so derivative of xy according to you is x'y+y'x= (x is constant y is constant) =0+0=0 $\endgroup$ Dec 15, 2013 at 16:09
  • $\begingroup$ It depends what you are differenciating against. if both $x$ and $y$ are constants, then their product is also a constant, and the result would indeed be $0$. But if we consider for example : $\frac{\partial x^2}{\partial x}=\frac{\partial x\times x}{\partial x}=\frac{\partial x}{\partial x}\times x +x \times \frac{\partial x}{\partial x}= 1\times x + x\times 1=2x$. If you are differenciating $xy$ with respect to $x$ (and $y$ is not a function of $x$), then $\frac{\partial x}{\partial x}=1$ and by using my formula $\frac{\partial xy}{\partial x}=y$ $\endgroup$
    – imj
    Dec 15, 2013 at 16:33

3 Answers 3

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Yes , He used simple method of derivative and differentiate as usual like df(x)/dx and assuming y as constant k .

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You answer is correct. When we want to take the partial derivative of a function with respect to one variable we consider all the other variables as a constants.

We also avoid to use $f^\prime$ notation because we don't know which variable you are taking derivatives if you use the prime notation. We use either $f_x$ or $\frac{\partial}{\partial x}f$.

A good exercise for you is to compute $\frac{\partial}{\partial y}z$. Can you show that $z_y=0+x+2y$ ?

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  • $\begingroup$ Yes because $x²+xy+y²$ is symmetric and we will now treat x as a constant and we get same result partialz\partialx but we change x with y and y with x. $\endgroup$ Dec 15, 2013 at 16:10
  • $\begingroup$ Yes, now practice with some non-symmetric examples if you have a chance. After solving more examples you will be comfortable and computing partial derivatives will be easy as taking regular derivatives $\endgroup$ Dec 15, 2013 at 16:32
  • $\begingroup$ Thank you. That helps me a lot $\endgroup$ Dec 15, 2013 at 16:34
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If f(x) refers to z, and f'(x) refers to $\frac{\partial z}{\partial x}$, what you did is right. You treat the y's as constants and differentiate with respect to x.

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