I know the definition of the partial derivative but there is another way for calculating the partial derivative by holding one variable constant and taking ordinary derivative . When this method works ? I mean when we can use this method instead of the definition and when we should apply the definition ?
For example :
$f(x,y) =
\begin{cases}
\frac{x^3 + x^4 – y^3}{x^2+y^2} & \text{ $(x,y) \not = (0,0)$ } \\
0 & \text{$(x,y) = (0,0)$}
\end{cases}$
We want to calculate $\frac{\partial f}{\partial x}(0 , 0)$ . Is it necessary to use definition here ?
Best Answer
Why are we allowed to find partial derivatives by holding a variable constant?
This can be seen directly from the definition:
$$\frac{\partial f(x,y)}{\partial x} = \lim\limits_{h\to0} \frac{f(x+h,y)-f(x,y)}{h}$$
In order to compute the limit, we can assume $y$ to be a parameter inside the limit. To make this more intuitive, we are going to let $g(x)=f(x,y)$. Applying this, we get
$$\frac{\partial f(x,y)}{\partial x} = \lim\limits_{h\to0} \frac{g(x+h)-g(x)}{h}$$
and this is just the derivative of a one-variable function. Additionally, when you are looking for derivatives the "easy way", you use various rules. But these rules sometimes come with some caveats, that we usually forget about, because most problems we solve are "nice enough" and do not require any special thought.
Why can you not use the method?
In this specific example, you would use the quotient rule, but, as any rule, it is only valid when the function is defined. This is clearly not the case for $(x,y)=(0,0)$ because the denominator is zero. So, the partial derivative of $\frac{x^3 + x^4 - y^3}{x^2+y^2}$ is undefined at $(0,0)$, and you cannot use the derivative of this function to find the derivative of $f$. But, this is not enough to say that $f$ has no partial derivatives at $(0,0)$, because it is defined there.
But, even if $f$ were defined at $(0,0)$, you would still be unable to use the method. Consider the following function:
$$f(x,y) = \begin{cases} x, & x\ne0 \\ 1, & x=0 \end{cases}$$
The derivative of $x$ is always $1$, but the partial derivative of $f$ with respect to $y$ doesn't exist at $0$. An easy way to see this is to notice that the function is not continuous at that point.