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I am learning partial derivatives. I am stuck in understanding the concept. I am expressing the question in form of problems below:

Problem1: $z = xy$, $x$ & $y$ are independent of each other.

what is change in $z$, $\Delta z$, at $x = 1, y = 1$, due to small change in $x$, $\Delta x = 0.001$ and a small change in $y$, $\Delta y = 0.001$

Problem2: $z = xy$

Also $y = x$ (i,e $y$ is function of $x$) what is change in $z$, $\Delta z$, at $x = 1, y = 1$ due to small change in $x$, $\Delta x = 0.001$ and a small change in $y$, $\Delta y = 0.001$

Thanks in advance for answers.

I found the explanation here http://en.wikipedia.org/wiki/Total_derivative#Differentiation_with_indirect_dependencies thanks 5xum

user642796
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Prashanth
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1 Answers1

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For a function $f$ of two variables, you can use the estimate $$f(x+\Delta x,y+\Delta y)\approx f(x,y) + \left(\frac{\partial f}{\partial x}(x,y)\right)\cdot\Delta x + \left(\frac{\partial f}{\partial y}(x,y)\right)\cdot \Delta y.$$

For a function $f$ of one variable, that simplifies to $$f(x+\Delta x)\approx f(x) + f'(x)\Delta x$$

5xum
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  • Yes using the above formula I get the answer for problem 1 as 0.002 – Prashanth Apr 29 '14 at 08:24
  • Now in Problem 2: I can take it as z = x^2, since y = x solving I will get change in z as 0.002 Is this correct? – Prashanth Apr 29 '14 at 08:33
  • Correct. In problem one, you get $\Delta z = \Delta x + \Delta y = 0.002$ as the partial derivatives equal $1$. In problem two, the function is $x^2$ so the derivative is equal to $2$ and $\Delta z = 2\Delta x = 0.002$ – 5xum Apr 29 '14 at 09:48
  • please edit your comment with dollar signs, it is unreadable at the moment. – 5xum Apr 29 '14 at 10:18
  • But i can also solve Problem2 this way: $$\Delta z = \left ( \frac{\partial z}{\partial x} \right )\Delta x +\left ( \frac{\partial z}{\partial y} \right )\Delta y$$

    Now $$z = x^2$$ so,
    $$\left ( \frac{\partial z}{\partial x} \right )\Delta x = \left ( 2x \right )\Delta x = 0.002$$

    and
    $$\left ( \frac{\partial z}{\partial y} \right )\Delta y = x\Delta y=0.001 $$

    So $$\Delta z = 0.003$$

     – Prashanth Apr 29 '14 at 10:25
  • No. If $z=x^2$, then $\partial z/\partial y = 0.$ – 5xum Apr 29 '14 at 10:26
  • But $$\partial z/\partial y \neq 0$$ How come making y=x make it $$\partial z/\partial y=0$$ Still z = xy, so changing y should change z – Prashanth Apr 29 '14 at 10:47
  • For $z=x^2$, you have $\partial z/\partial x = 2x$ and $\partial z/\partial y = 0$. For $z=xy$, you have $\partial z\partial x=y$ and $\partial z/\partial y = x$. You can't take one derivative from $z=x^2$ and the other from $z=xy$. – 5xum Apr 29 '14 at 10:59
  • Lets say there is one small circular metal disc On side it is heated. Now since it is heated, it can absorb some energy, its temperature increases and its radius increases. – Prashanth Apr 30 '14 at 05:17
  • Also we can increase its radius directly by stretching

    Now lets say the metal disc release the heat on other side

    This energy released, depends on its temperature & its surface area.

    Now at some instant the heating temperature is increased by small amount and also its radius is increased by streatching. Now we have to find the increase in energy released by this metal disc.

    Shouldn't we consider the increase in radius two times? One due to increase in temperature of plate & another due to increase in streatching?

     – Prashanth Apr 30 '14 at 05:17
  • Yes, and that is why you have $2\Delta x$ for $z=x^2$. The $2$ comes from your explanation. – 5xum Apr 30 '14 at 06:31