160

The reason is the following. We use the notation: $$\theta x^i:=\theta_0+\theta_1 x^i_1+\dots+\theta_p x^i_p.$$ Then $$\log h_\theta(x^i)=\log\frac{1}{1+e^{-\theta x^i} }=-\log ( 1+e^{-\theta x^i} ),$$ $$\log(1- h_\theta(x^i))=\log(1-\frac{1}{1+e^{-\theta x^i} })=\log (e^{-\theta x^i} )-\log ( 1+e^{-\theta x^i} )=-\theta x^i-\log ( 1+e^{-\theta x^i} ),$$ [...


40

Congratulations, you have met one of the worst ambiguities in mathematical notation! Assume you have a function of two variables, $f \colon A \times B \to \mathbb{R}$, where $A$ and $B$ are subsets of $\mathbb{R}$. The notation $$\frac{\partial f}{\partial x}(x_0,y_0)$$ is commonly used to denote the value of the partial derivative of $f$ with respect to ...


33

Despite the popularity of the top answer, it has some major errors. The most fundamental problem is that $g(f^{(i)}(\theta_0, \theta_1))$ isn't even defined, much less equal to the original function. The focus on the chain rule as a crucial component is correct, but the actual derivation is not right at all. So I'll give a correct derivation, followed by my ...


32

Let $f(x,y)$ be some function. Then you can visualize $\partial f / \partial x$ evaluated at a given $y=y_0$ as the derivative of a new function $g(x) = f(x, y_0)$. In particular, $\partial f / \partial x \rvert_{y=y_0}=dg/dx$. Geometrically, visualize the graph of $f(x,y)$: some surface. Take a plane that is parallel to the $x$ and $z$ axes and slice the ...


32

$$\frac{dg(x)}{df(x)} = \frac{dg(x)}{dx} \cdot \frac{1}{f'(x)} = \frac{g'(x)}{f'(x)}$$ In your example, $$g'(x) = 2f'(x) + 1 + \frac{f'(x)}{f(x)}$$ So: $$\frac{dg(x)}{df(x)} = \frac{2f'(x) + 1 + \frac{f'(x)}{f(x)}}{f'(x)} = 2 + \frac{1}{f'(x)} + \frac{1}{f(x)}$$


29

If we define the derivative of $f(z)$ as the limit as $h$ approaches $0$ (being $h$ a complex number) of $(f(z+h)−f(z))/h$... That's precisely what we do. then clearly the derivative is $2z$, but what does this derivative represent?? Well, it represents what we have defined: the limit of the incremental ratio, same as in the real case. Probably you're ...


26

Some key things to remember about partial derivatives are: You need to have a function of one or more variables. You need to be very clear about what that function is. You can only take partial derivatives of that function with respect to each of the variables it is a function of. So for your Example 1, $z = xa + x$, if what you mean by this to define $z$ ...


25

It means that $z$ is a function of $x$ and $y$, so the full writing would be $$x^2+y^2+(z(x,y)-c)^2=a^2$$


21

You can do it componentwisely: $$\frac{d(x^Ta)}{dx}=\left(\frac{d(x_1a_1+x_2a_2+\cdots+x_na_n)}{dx_1}, \frac{d(x_1a_1+x_2a_2+\cdots+x_na_n)}{dx_2}, \cdots, \frac{d(x_1a_1+x_2a_2+\cdots+x_na_n)}{dx_n}\right)\\ =(a_1,a_2,\cdots, a_n)=a^T$$


21

You may interchange integration and differentiation precisely when Leibniz says you may. In your notation, for Riemann integrals: when $f$ and $\frac{\partial f(x,t)}{\partial x}$ are continuous in $x$ and $t$ (both) in an open neighborhood of $\{x\} \times [a,b]$. There is a similar statement for Lebesgue integrals.


20

When you take the derivative of $\frac{y}{x}$ with respect to $y$ you are computing $\frac{\partial }{\partial y} \frac{y}{x} = \frac{1}{x}$ because here you are holding $x$ constant. If you take the derivative of the same expression with respect to $x$ then you compute $\frac{\partial}{\partial x} \frac{y}{x} = - \frac{y}{x^2}$ and this is when you hold $y$...


15

To understand some formulas from calculus intuitively, we could try to interpret the notions in discrete ways. For $y\in [c,d]$, the values $$F(y) = \int_a^b f(x,y)dx$$ denote the areas of the slices (integrals along the x axis) specified by $y$: To measure the difference between each two adjacent areas (discrete analog of $F'(y)$), one tried to draw all ...


14

That matrix is symmetric. It is a consequence of linear algebra that a symmetric matrix is orthogonally diagonalizable. That means there are two perpendicular directions upon which that matrix acts as scaling by $\lambda_1$ and by $\lambda_2$. These $\lambda_i$ represent the quadratic coefficient of a parabolic approximation to the function $f$ at $(x_0,y_0)...


14

More generally, if $u(x_1,\ldots,x_n)$ is a partially differentiable function function in $n$ variables and $s_1,\ldots ,s_n$ are differentiable and $f(t)=u(s_1(t),\ldots,s_n(t))$ then $$\frac {df}{dt}=\frac{\partial u}{\partial x_1} \frac {d s_1}{d t}+\ldots +\frac{\partial u}{\partial x_n} \frac {d s_n}{d t}$$ Your $(\mathrm{A})$ is a special case of $n=2$...


14

In Dieudonné's Elements of Analysis, volume 1, you'll find an exercise which states that if $f$ is $C^1$ and $f_{xy}$ exists and is continuous on an open set $U$, then $f_{yx}$ exists and equals $f_{xy}$ on $U$. (So, of course, $f_{xy}$ is continuous as well.) Here's a sketch of the proof. As in the usual proof that a $C^2$ function has equal mixed partial ...


14

The notation $\dfrac {\partial}{\partial x}$ indicates that all variables other than $x$ should be treated as constant, whereas $\dfrac{d}{dx}$ would treat the other variables as exactly that: variable. Thus for $f(x,y)=yx^3$, we have $$ \begin{align} \frac{\partial f}{\partial x}&=3yx^2 \\ \frac {df}{dx}&=3yx^2+ \frac {dy}{dx}x^3 \\ \end{align} $$


14

A complex-valued function $f$ is differentiable if near an arbitrary point $z_{0}$, the function $f$ acts like multiplication by a complex number $f'(z_{0})$ in the sense that $$ f(z_{0} + h) = f(z_{0}) + h\, f'(z_{0}) + o(|h|). $$ Since multiplication by a non-zero complex number $$ \alpha = |\alpha| \exp(i\theta) $$ rotates the plane (about the origin) ...


14

When god created the world ${\mathbb R}^3$ the three variables $x$, $y$, $z$ used to address the points $p\in{\mathbb R}^3$ were truly independent. Now an equation $$x^2+y^2+(z-c)^2=a^2\tag{1}$$ is given, whereby for simplicity we assume $a>0$. Such an equation does not define any function per se, but a solution set $$S:=\bigl\{(x,y,z)\,\bigm|\,x^2+y^2+(...


13

By first principles, $ \dfrac{\partial}{\partial y}\left( \dfrac{y}{x}\right)=\lim\limits_{\delta y\to 0}\left( \dfrac{\dfrac{y+\delta y}{x}-\dfrac{y}{x}}{\delta y}\right)=\lim\limits_{\delta y\to 0}\dfrac{\delta y}{x\delta y}=\lim\limits_{\delta y\to 0}\dfrac{1}{x}=\dfrac{1}{x} $


13

I hope this answers your question. The partial derivative notation is used to specify the derivative of a function of more than one variable with respect to one of its variables. e.g. Let y be a function of 3 variables such that $y(s, t, r) = r^2 - srt$ $$\frac{\partial y}{\partial r} = 2r-st$$ $\frac{d}{dx}$ notation is used when the function to be ...


13

Let $w=f(z) = z^2$. Then $f(2+i)= 3+4i$ and $\dfrac{dw}{dz} = f'(z) = 2z$ and $f'(2+i) = 4+2i$. The means an infinitely small change in $z$, from $2+i$ to $2+i+dz$, has a corresponding infinitely small change in $w$, from $3+4i$ to $3+4i + dw = 3+4i + (4+2i)\,dz$. The change $dz$ gets multiplied by $4+2i$. Geometrically, what is a multiplication by $4+2i$...


12

First, rest assured that you're not the only one who's confused by the standard notation for partial derivatives. See this answer for a collection of answers I've written in response to such confusions. The problem is that the standard notation doesn't indicate which variables are being held constant. It assumes that you've defined a function of a certain ...


12

When you write $\displaystyle \frac{dz}{dx}$ you are assuming that $z$ is a function of $x$ only. This forces $y$ to be a function of $x$, therefore abusing the notation we write $y=y(x)$. Meanwhile, we have that $f$ continues to be a function of two variables, therefore $\displaystyle \frac{\partial f}{\partial x}$ means the usual partial derivative with ...


11

Rewrite your formula as $$\frac{\partial}{\partial x} = \cos\theta\frac{\partial}{\partial r}-\sin\theta \frac{1}{r}\frac{\partial}{\partial\theta}\tag{1}$$ (I like writing in this way because $\frac{1}{r}\frac{\partial}{\partial\theta}$ is more natural than $\frac{\partial}{\partial\theta}$: it means the rate of change in the tangential direction). Now ...


11

I don't understand what you say about derivatives and I will assume that you want the following result. Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function such that for every $a, b \in \mathbb{R}$, $x \mapsto f(x,b)$ and $y \mapsto f(a,y)$ are polynomial functions; then $f$ is a polynomial in two variables. This is not obvious because we don't ...


11

If we identify the functions of $ \ x \ $ and $ \ y \ $ involved in the definition of the function $ \ \varphi \ $ by $$f(x,y) \ = \ \varphi (\ \underbrace{\frac yx}_u \ , \ \underbrace{ x^2-y^2}_v \ , \ \underbrace{y-x}_w \ ) \ , $$ we can use the multivariate extension of the Chain Rule to write $$ \frac{\partial f}{\partial x} \ = \ \frac{\partial \...


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