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I have some questions about the definitions of the derivatives of real valued functions which are mostly to make sure I got things correctly. Are the following statements correct?

The analogous of partial derivatives of functions $f:\Bbb R^n \to \Bbb R$ for functions $g:\Bbb R \to \Bbb R$ is derivatives, correct? The analogous of the derivative of functions such as $f$ for functions such as $g$ can also be defined and it is something different than the usual derivative of $g$, correct?

(If $g(x)=(x-2)^3+3$ then $g'(x)=3(x-2)^2$ which is not a linear map but $Dg=$??)

If the above are correct then we call these linear mappings derivatives because they simply have very similar properties to the derivatives of functions defined in $\Bbb R$.

I hope an answer alleviates this confusion. Thanks in advance

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If $g(x) = (x-2)^3$, then $g'(x)=3(x-2)^2$, which is not a linear function of $x$, that's correct. What's linear is $\Delta g \approx g'(x)\cdot h$, the change in the tangent line as a function of $h$, the change in $x$.

More generally in multivariable calculus, the Jacobian $J(f) = \frac{\partial f(x,y)}{\partial(x,y)}$ is a matrix which may depend in a nonlinear way on $x$ and $y$, but multiplies linearly with displacement vectors in $x$ and $y$.

Or more succinctly, $df_p(v)$ is linear in $v$, not in $p$.

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Given a function $f:\>{\mathbb R}^n\to{\mathbb R}^m$ and a point $p$ in the domain of $f$ the derivative of $f$ at $p$, if it exists, is a linear map $df(p):\>T_p\to T_{f(p)}$ characterized by $$f(p+X)-f(p)=df(p).X+o\bigl(|X|\bigr)\qquad(X\to0)\ .\tag{1}$$ (Here $T_p$ is the tangent space at $p$, a copy of ${\mathbb R}^n$ with origin at $p$.) This general abstract concept makes sense also in the special cases $m=1$ (a scalar-valued function $f$) or $n=1$ (a function $f$ of one variable). In the Calculus 101 case $f:\>{\mathbb R}\to{\mathbb R}$ formula $(1)$ reads $$f(p+h)-f(p)=f'(p) h+o\bigl(|h|\bigr)\qquad(h\to0)\ ,\tag{2}$$ whereby I have written the increment vector $X$ as $h$.

Note that only the value of the derivative $df$ at the point $p$, the single number $f'(p)$, enters formula $(2)$. You got mixed up because hearing "derivative" you think of a function $x\mapsto f'(x)$, whereas in the above discussion a single "working point" $p$ is at stake. The multivariate analogon to the derivative function $x\mapsto f'(x)$ is the matrix-valued function $x\mapsto df(x)$, which does in no way depend linearly on $x$.

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  • $\begingroup$ In the case when m=1 the function $df(a)(h)=\frac{\partial f}{\partial x_1}(a)h_1+...+\frac{\partial f}{\partial x_n}h_n$. So in the case when $n=1$ df(a)(x)=\frac{\partial f}{\partial x}(a)x=\frac{df}{df}(a)x, which passes through the origin..so this doesn't seem to me that it is the function $x \to f'(x)$ $\endgroup$ – Zero Pancakes Nov 8 '17 at 15:55

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