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I am bit confused regarding the geometrical/logical meaning of partial and total derivative. I have given my confusion with examples as follows

Question

  1. Suppose we have a function $f(x,y)$ , then how do we write the limit method of representing $ \frac{\partial f(x,y)}{\partial x} \text{and} \frac{\mathrm{d} f(x,y)}{\mathrm{d} x}$ at (a,b)? What is the difference?

  2. Imagine I have a function $f(t,x,y)= t^4+x^2+y^3+2xy^9\tag 1$

    what could be $ \frac{\partial f(t,x,y)}{\partial x} , \frac{\mathrm{d} f(t,x,y)}{\mathrm{d} x},\frac{\partial f(t,x,y)}{\partial t} , \frac{\mathrm{d} f(t,x,y)}{\mathrm{d} t}$? What is the difference between them in meaning?

  3. Imagine I have a function $f(t,x,y)= t^4+x^2+y^3+2xy^9,x=\psi(t),y=\tau(t)\tag 1$

    what could be $ \frac{\partial f(t,x,y)}{\partial x} , \frac{\mathrm{d} f(t,x,y)}{\mathrm{d} x},\frac{\partial f(t,x,y)}{\partial t} , \frac{\mathrm{d} f(t,x,y)}{\mathrm{d} t}$? What is the difference between them in meaning?

  4. When we can say $ \frac{\partial f(x,y)}{\partial x} = \frac{\mathrm{d} f(x,y)}{\mathrm{d} x}$? Can we say $\frac{\partial^2 f(t,x,y)}{\partial x dy} =\frac{\partial^2 f(t,x,y)}{ dy \partial x}$ if function $f(t,x,y)$ is $C^2$ continuous?

  5. We have a function curve ${f(t,g_1,g_2,g_3)}_{3 \times 1} \tag 2$

    and we have a $4 \times 1$ array called $p=\begin{pmatrix}t\\ g_1\\ g_2\\ g_3 \end{pmatrix}$ All t and $g_i$ are variables, t is the curvilinear parameter . What is the meaning difference between $\frac{\partial f(t,g_1,g_2,g_3)}{\partial p} \text{and} \frac{d f(t,g_1,g_2,g_3)}{d p} $? How do we express the difference in limit notations?

NB : Replies written with question enumeration will be more helpul. These are basics, I know. But I get confused some time. Looking for interpretations beyond equations. Means geometrically or logically coherent one.This is not a homework problem just because it is simple. I made the example for expressing my issue so that I can learn from the result. Thanks for the time to read my question.

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    $\begingroup$ Is this a homework assignment? $\endgroup$ – Paul Sundheim Sep 5 '14 at 17:02
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    $\begingroup$ No No I just made example for expressing my problem.. You can see I am asking meaning than solution $\endgroup$ – Sri Hari Sep 5 '14 at 17:23
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As Paul said, the formal definition of a partial derivative is the limit $$\frac{\partial f}{\partial x}(a,b)=\lim\limits_{h \rightarrow 0} \frac{f(a+h,b)-f(a,b)}{h}.$$

This definition must be interpreted as a slope of the tangent line at the point $(a,b)$ for the intersection curve between the plane $y=b$ and the surface $f(x,y)$.

Now, for functions of more than one independent variable $f(x_1,x_2,...,x_n)$ the sum of the partial differentials with respect to all of the independent variables is the total differential $$df=\frac{\partial f}{\partial x_1}dx_1+\frac{\partial f}{\partial x_2}dx_2+ \cdots +\frac{\partial f}{\partial x_n}dx_n=\sum\limits_{i=1}^{n}\frac{\partial f}{\partial x_i}dx_i.$$

Suppose that $f(t,x,y)$ with $x=\psi(t), \, y=\tau(t)$. Then we have $$\frac{df}{dt}=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt},$$ since $\frac{dt}{dt}=1$ and $x'(t)=\frac{dx}{dt}$ because the function $x(t)$ is a real valued function of a real variable

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  1. $\frac{\partial f}{\partial x}(a,b)=lim_{h\to 0}\frac{f(a+h,b)-f(a,b)}{h}$. $\frac{d f}{d x}(a,b)$ has no meaning, unless $y$ is a function of $x$. I will address the other questions as I get some time.
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