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I'm having some difficulty understanding the meaning of "hold y (or x) constant" when finding the partial derivative of an equation such as:

$z=x^2+xy+y^2$

Please show all intermediate steps using partial derivative notation and explain what it means in practical terms to hold a variable constant using actual numbers.

Thanks!

user913304
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1 Answers1

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One way of understanding it is to introduce a one-variable function and then differentiate that function in the way you already know. For instance, if you are asked to compute $\frac{\partial z}{\partial x}$ at $(x,y)=(1,2)$, you can introduce $r(x)=z(x,2)=x^2+2x+4$. Then $\frac{\partial z}{\partial x}(1,2)=r'(1)$.

If you want to get $\frac{\partial z}{\partial x}$ for general $(x,y)$, you can imagine having a different function $r$ for each value of $y$, and differentiating that function. This is not a very convenient way to do calculations, but it is a simple way of thinking about what's going on.

As a practical matter, when you compute $\frac{\partial z}{\partial x}$, you treat $y$ as if it were a number. Thus for instance $\frac{\partial}{\partial x}(xy)=y$, for the same reason that $\frac{d}{dx}(2x)=2$. Similarly $\frac{\partial}{\partial x}(y^2)=0$, for the same reason that $\frac{d}{dx}(4)=0$.

Ian
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  • So, the example in my question would become:

    $\frac{∂z}{∂x}(x^2+xy+y^2)=\frac{d}{dx}(x^2)+\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)$

    And, if we work those out individually:

    $\frac{d}{dx}(x^2)=2x$

    (x term, but no y term, therefore: differentiate with respect to x, $x^2$ = 2x)

    $\frac{d}{dx}(xy)=y$

    (x term and a y term, therefore: differentiate with respect to x, x = 1 and y remains untouched, as it is held constant)

    $\frac{d}{dx}(y^2)=0$

    (y term but no x term, therefore: differentiate with respect to x, $y^2$ = 0)

    Yielding:

    $\frac{∂z}{∂x}(x^2+xy+y^2)=2x+y$

    Is this correct?

     – user913304 Oct 20 '15 at 23:04
  • @user913304 Yup. – Ian Oct 20 '15 at 23:06