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Suppose that $f:\mathbb{R}\to {\mathbb R}^n$ is a smooth function with value $f(t)$ at $t$. What is the difference between $$\frac{df}{dt}$$ and $$\frac{\partial f}{\partial t}?$$ When partial and total differentiation are differenet?

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    $\begingroup$ How can they be different, if you only have one variable? $\endgroup$ – 57Jimmy Jan 10 '18 at 15:26
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A partial derivative ($\frac{\partial f}{\partial t}$) of a multivariable function of several variables is its derivative with respect to one of those variables, with the others held constant.

Let $f(t,x)=t^2 + tx + x^2$. Then $$\frac{\partial f}{\partial t} = 2t + x + 0$$

On the other hand, the total derivative ($\frac{\mathrm df}{\mathrm d t}$) is taken with the assumption that all variables are allowed to vary. So

$$\frac{\mathrm df}{\mathrm d t} =\frac{\partial f}{\partial t}\frac {\mathrm d t}{\mathrm dt} + \frac{\partial f}{\partial x}\frac {\mathrm d x}{\mathrm dt}$$

Hence if $f(t,x)=t^2 + tx + x^2$ and $\frac {\mathrm d x}{\mathrm dt}=t^3$, then $$\frac{\mathrm df}{\mathrm d t} =(2t +x) + (2x+t)(t^3)$$

Of course, if the function is a function of only one variable, then the total and partial derivatives are the same.

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Partial differentiation is used when the domain of the function is defined on a cartesian product set. Supposed that $f : E \times F \to F$ with $(x,y) \mapsto f(x,y)$ then you have two partial differentiation to separate the derivative against the first variable namely $\frac{\partial f}{\partial x}$ vs. the second variable $\frac{\partial f}{\partial y}$. If the domain isn't a cartesian product (like for your example), then the partial and total differentiation coincide.

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