New answers tagged

1

$$f'(x)=\lim_{h \to 0} \frac{h \sin (\pi/h)-0}{h}= \lim_{h \to 0} \sin (\pi/h)=\text{does not exist}.$$ $L=\lim_{x \to 0} \sin(\pi/x)$ does not exist for two sequeces $x_n=\frac{1}{n}$ and $x'_n=\frac{1}{(n+1/2)}$, let $g(x)=\sin(\pi/x)$, then $g(x_n)=0, g(x'_n)=1$ these being different there are two accumulation points. So, the limit $L$ does not exist. ...


1

The ODE: $$(xln(y) + xy)dx + (yln(x) + xy)dy=0~~~~(1) $$ can be re written as $$\int \frac{xdx}{x+\ln x}+\int \frac{ydy}{y+\ln y}=c~~~~(2)$$ by dividing (1) by $(x+\ln x)(y+\ln y)$ (2) is separable yest the integrals are not doable. So $[(x+\ln x)(y+\ln y)]^{-1}$ can be taken as the integrating factor for (1).


3

Both coefficient terms are separable, so that you can separate terms in $x$ and $y$. $$ \frac{x\,dx}{\ln x+x}+\frac{y\,dy}{\ln y+y}=0. $$


1

The curve is indeed not the graph of a function. At any point $(x,y)$ on the curve, if an open disk about that point is small enough, then that portion of the curve that is within that neighborhood is the graph of a function, and the slope of the tangent line to the graph of that function is $-x/y.$ Derivatives are local, that is the slope of a curve at a ...


1

You can still think of it in terms of the slope of a tangent line, and even in terms of a limit. However, as you've pointed out, $x^2 + y^2 = r^2$ isn't a function because it fails the vertical line test. However, by the Implicit Function Theorem we can consider $F(x,y) = x^2 + y^2 - r^2$, and for any $(x_{0},y_{0})$ where $\frac{\partial F}{\partial y}\ne ...


7

From the mean-value theorem you get $$ \frac{f(x)-f(0)}{x-0} > 2 $$ for all nonzero $x$. It follows that $$ f(x) \ge f(0) + 2x \text{ for } x > 0 $$ and $$ f(x) \le f(0) + 2x \text{ for } x < 0 $$ so that $f$ necessarily takes both positive and negative values. Now conclude with the intermediate value theorem.


1

While each $x\in(-r,\,r)$ is compatible with two choices of $y$, continuous motion along the circumference well-defines the choice of $y$ at each point, giving a local $y$-as-a-function-of-$x$ behaviour wherever $dy/dx$ is finite and nonzero (i.e. $x,\,y$ are both nonzero), which happens at all but four of the circumference's points. This local behaviour is ...


0

$\pi_1(t|v_1)=b^*(t)^{n-1}(v_1-b^*(t))$ As $\pi_1(t|v_1) = \color{red}{b^*(t)^{n-1}}\times \color{blue}{(v_1-b^*(t))}$ so according the product rule $\frac {d\pi_1'(t|v_1)}{dt} = \frac {d\color{red}{b^*(t)^{n-1}}}{dt}[\color{blue}{(v_1-b^*(t))}] + [\color{red}{b^*(t)^{n-1}}]\frac {d\color{blue}{(v_1-b^*(t))}}{dt}$ $\color{red}{b^*(t)^{n-1}} = \color{green}(\...


1

This is hint in the form of a suggestion, as one way to approach the problem. Since you don't have to derive the expression for $h(t)$, you could first check that it satisfies the integral equation, and second show that the integral equation has a unique solution. One way to show uniqueness would be to derive from the integral equation, an ordinary ...


2

This is only a broad outline and not a detailed answer. $f$ has right-hand and left-hand derivatives at every point. These are both monotone functions and hence they are continuous except at countable number of points. At any point where one the derivatives is continuous the function is actually differentiable. Hence $f$ is differentiable at all but ...


1

Well, if $i \neq j$ then $\frac{ \partial \sigma(x_i)}{\partial x_j} = 0 $ so $\nabla f$ is sparse, only the n entries along the diagonal are non-zero, etc.


1

Check that $$\int_0^x (x-t)^2 \sin^2\frac{t}{2} d\,t= \sin x - (x - \frac{x^3}{6})$$ WA link


2

To show it is constant, show that the derivative is $=0$. $$\begin{align}\frac {\mathrm d}{\mathrm dx}\sum_{k=0}^n(-1)^{n-k}f^{(n-k)}g^{(k)}&=\sum_{k=0}^n(-1)^{n-k}(f^{(n-k+1)}g^{(k)}+f^{(n-k)}g^{(k+1)})\\ &=\sum_{k=0}^n(-1)^{n-k}f^{(n+1-k)}g^{(k)}-\sum_{k=1}^{n+1}(-1)^{n-k}f^{(n+1-k)}g^{(k)}\\ &=(-1)^nf^{(n+1)}g^ {(0)}-(-1)^{-1}f^{(0)}g^{(n+1)}\\...


2

You are right. Left & right derivatives and directional derivative are different notions – they coexist in fact, also in one-dimensional spaces such as $\mathbf{R}$. The directional derivative can be defined on a generic vector- or affine space or on a differential manifold, without the existence of a norm or scalar product. It's therefore useful not ...


1

The main feature of the Leibniz law $f(x)g(x)' = f'(x)g(x) + f(x)g'(x)$ is that it turns a product into a sum. Another way to write it to show that more explicitly is $$ \frac{d}{dx}\left(f(x)g(x)\right)=\left(\left.\frac{\partial}{\partial s}\right\rvert_{s,t=x}+\left.\frac{\partial}{\partial t}\right\rvert_{s,t=x}\right)\left(f(s)g(t)\right)=\frac{df}{dx}g(...


2

Anyway, no need for L'Hospital's rule here. Just use Maclaurin's expansions at the relevant orders and compose them where required: $\mathrm e^x=1+x+o(x)$; $\sin^2x=x^2+o(x^2)$, so $\mathrm e^{\sin^2x}=1+x^2+o(x^2)=1+o(x)$; $\sin 2x=2x+o(x)$. Therefore $$\frac{\mathrm e^{\sin^2x}-\mathrm e^x}{ {\sin2x}}=\frac{-x+o(x)}{2x+o(x)}=-\frac12+o(1).$$


5

Just to give an alternative to J.G.'s approach, we have $$\begin{align} {e^{\sin^2x}-e^x\over\sin2x} &={(e^{\sin^2x}-1)-(e^x-1)\over2\sin x\cos x}\\ &={1\over2\cos x}\left({e^{\sin^2x}-1\over\sin^2x}\sin x-{e^x-1\over x}\cdot{x\over\sin x} \right)\\ &\to{1\over2\cdot1}\left(1\cdot0-1\cdot1 \right)\\ &=-{1\over2} \end{align}$$


0

First of all, the standard notation for the second (and higher) derivatives is misleading, because it is non-algebraic (i.e., the ratios don't cancel). It's even more problematic when using partial differentials, because the symbols themselves are equivocal within the same expression. In any case, what happens when you take a derivative is that you first ...


9

You want$$\underbrace{\lim_{x\to0}e^x}_{1}\cdot\underbrace{\lim_{x\to0}\frac{e^{\sin^2x-x}-1}{\sin^2x-x}}_{1}\cdot\lim_{x\to0}\frac{\sin^2x-x}{\sin(2x)}$$(the second limit uses $\lim_{y\to0}\frac{e^y-1}{y}=1$). The last limit is$$\underbrace{\lim_{x\to0}\frac12\tan x}_0-\underbrace{\lim_{x\to0}\frac{x}{\sin(2x)}}_{1/2}=-\frac12.$$


1

You can simply use the equivalence of the functions sinus and exponential at $x=0$. Thus $\sin(x)$ is equivalent to $x$ at $x=0$, and $(\sin(x))^2$ is equivalent to $x^2$ at $x=0$. In addition, $\exp(x)$ is equivalent to $1+x$ at $x=0$. Bearing this in mind, you find that $\exp( (\sin(x))^2 ) - \exp(x)$ is equivalent to $(1+x^2) - (1+x) = x(x-1)$ at $x=0$. ...


2

Hint: You have $$\frac{1}{f(f-1)}\,f'=1,$$ which becomes $$\int\frac1{f(f-1)}\,df=\int1\,dx.$$


0

The directional derivative is not the same as the limit approaching from the left or the right. If you use the negative unit vector, you're asking what the slope is if you're traveling to the right vs. to the left. You get the negative of the slope if you use the negative unit vector. But that is correct. The divide by the magnitude of v ($|v|$) is ...


2

Hint: $$f^\prime = f (f-1)$$ Is a separable differential equation. It's also A Bernoulli's differential equation: $$f'+f= f^2$$ $$-\left (\dfrac 1 f \right)'+\dfrac 1 f=1$$ $$u'-u=-1$$ Then use integrating factor: $$(ue^{-x})'=-e^{-x}$$ Inetgrate and unsibstitute $u=\dfrac 1f$.


1

What was the original statement of the problem? If we interpret $u(t)$ as a vector, then it's not a single vector, but rather an infinite family of different vectors — one for each real value of $t$. And we can find the corresponding family of answers, but I doubt that's what this question means. It looks more likely that the given $u(t)$ represents a line, ...


2

You have $$\dfrac{\sin (60+\alpha)}{\sin \alpha}=\dfrac{5}{4}$$ $$\dfrac{\sin 60 \cos \alpha+\cos 60 \sin \alpha}{\sin \alpha}=\dfrac{5}{4}$$ $$\Rightarrow \cot \alpha = \dfrac{\sqrt{3}}{2}$$ $$\Rightarrow \sin \alpha = \dfrac{1}{\sqrt{1+\cot ^2 \alpha}}=\dfrac{2}{\sqrt 7}$$ $$\therefore \sin (60+\alpha) = \dfrac{5}{4} \cdot \dfrac{2}{\sqrt 7}$$


1

The given solution is not correct - although its final result is right - because we do not know whether $f'$ exists around $2$ and $1$, resp. So, we should only use the existence of the two given derivatives and this is possible: We know $$\lim_{t\to 0}\frac{f(2+t)-f(2)}{t}=f'(2)=6 \text{ and } \lim_{t\to 0}\frac{f(1+t)-f(1)}{t}=f'(1)=4$$ Hence, only using ...


0

This is some version of the "l'Hospital" rule. Or some truncated version of Taylor expansion. I don't like it. What you can do is this : using Taylor expansion for function $f$ at point $x$, you have : $$f(y)=f(x)+(y-x)f'(x) + o(y-x)$$ So, putting $x=2$ and $y=2+2h+h^2$ : $$f(2h+2+h^2) = f(2) + (2h+h^2)f'(2)+o(h) = f(2)+h((2+h)f'(2)+o(1))$$ and ...


3

For the first part, just rewrite $n(f(x_0 + \frac{1}{n}) - f(x_0))$ as $$ \frac{f(x_0 + \frac{1}{n}) - f(x_0)}{x_0 + \frac{1}{n} - x_0} $$ and apply the definition of differentiability. For a counterexample to the converse part, note $g : \mathbb{R} \to \mathbb{R}$ $$ g(x) = \begin{cases} 1,\ x \geq 0\\ 0,\ x < 0 \end{cases} $$ satisfies $$ \lim_{n \to \...


2

$f$ is concave and positive, and therefore an increasing function. It follows that $L = \lim_{x \to \infty} f(x)$ exists as a finite value or $+ \infty$. $f'$ is decreasing, so that $A = \sum_{n=1}^\infty f'(n)$ converges if and only if $$ \int_1^\infty f'(x) \, dx = \lim_{x \to \infty} f(x) - f(1) < \infty \, , $$ i.e. if $L$ is finite. Also $$ \left( \...


0

Another counter example: $$\int_0^2{(x-1)}{dx} = \int_0^2{(1-x)}dx = 0$$ But: $$\int_0^2{(x-1)(1-x)}dx = -\int_0^2{(x^2+2x+1)}dx = -8/3 -4 - 2$$


2

For the first question, consider the case of $f(x) = g(x) = \sin(x)$. Then the integral from $0$ to $2\pi$ would be $$\int_0^{2\pi}\sin^2(x)dx = \pi$$ So it doesn't have to equal $0$, meaning that the first statement is not necessarily true. For the second question, let $f(x) = g(x) = \sin(x)+1$. Let $\phi(x) = \cos(x)-\sin(x)$. The sum of the integrals ...


0

Write the function as $$\frac{\left(\sqrt{x^2-x+1}-a x-b\right) \left(\sqrt{x^2-x+1}+a x+b\right)}{\sqrt{x^2-x+1}+a x+b}$$ Expand, simplify and collect $x$ $$\frac{-\left(a^2-1\right) x^2-x(2ab+1)-b^2+1}{\sqrt{x^2-x+1}+a x+b}$$ to give $0$ as limit it must be $$ \begin{cases} a^2-1=0&a=\pm 1\\ 2ab+1=0&b=-\frac{1}{2a}\\ \end{cases} $$ $a=1;\;b=-\...


0

There is another way to solve $$y'=y+x$$ Let $y=z-x$ to make $$z'-1=z\implies z'-z=1$$ Now, the homogeneous solution is $z=c e^{x}$. Use now the variation of parameter to get $$c' \,e^{x}=1 \implies c'=e^{-x}\implies c=-e^{-x}+k\implies z=-1+ke^{x} $$ Back to $y$ $$y=-x-1+ke^{x}$$ and using the condition $k=2$ $$y=2e^{x}-x-1$$


1

Yes, continuity at $c$ is an essential requirement. Otherwise, you can make $f$ discontinuous at $c$, which would imply $f$ is not differentiable at $c$. $$ f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c} $$ Since $f$ is continuous at $c$, $\lim_{x\to c}(f(x)-f(c))=\lim_{x\to c}(x-c)=0$, and you can apply L'Hospital's rule: $$ \lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\...


2

You went wrong right at the beginning (assuming that you wrote the problem correctly). $$H = g \circ g = g(g(x)) $$ Composition not multiplication. I'm assuming that you can get $g(3)= 4$ easily. So $$H' = g \circ g = g'(g(x))g'(x)$$ Using rule for derivate inverses. $$g'(3) = \frac{1}{f'(g(3))} = \frac{1}{f'(4)} = 2$$ $$g'(4) = \frac{1}{f'(g(4))} = \frac{1}{...


0

They are. When your functions are "nice" enough (in this case, for example, $C^3$ or more), third derivatives commute (with no change of any sign). And this easily follows from second derivatives' commutativity (Schwarz's theorem). For example: $$ \partial_j\partial_k f=\partial_k\partial_j f \;\Rightarrow\; \partial_i(\partial_j\partial_k f)=\...


1

I solve $y' = y + x, \; y(0) = 1 \tag 1$ in another "real way", as follows: write the equation as $y' - y = x; \tag 2$ make the clever observation that $(e^{-x}y)' = -e^{-x}y + e^{-x}y'$ $= e^{-x}y' - e^{-x}y = e^{-x}(y' - y); \tag 3$ in light of (2) this becomes $(e^{-x}y)' = xe^{-x}; \tag 4$ integrate 'twixt $0$ and any value of $x$: $e^{-x}y(x) ...


0

Note that $\lim_{x\to \pi/2}{\arctan\left(\frac{x}{\tan(x)-x}\right)}=0$. Then this is just the limit definition of the derivative: $$ \lim_{x\to \pi/2}\frac{\arctan\left(\frac{x}{\tan(x)-x}\right)-0}{x-\pi/2} $$ $$ = \lim_{x\to \pi/2}\frac{d}{dx}{\arctan\left(\frac{x}{\tan(x)-x}\right)} $$A tedious chain-rule calculation produces: $$ = \lim_{x\to \pi/2}\...


2

This is trivial... when you know the theory of differential equations. I will lead you to a solution by an ad-hoc trial and error process. You correctly observe that $(e^x)'=e^x$. If we plug this tentative solution in the equation, we have $$e^x\color{red}=x+e^x$$ which is not what we want because of the term $x$. So let us try $y=e^x-x$, and we get $$e^x-1\...


4

Here is a proof directly from the definition of the derivative. We have: $$\begin{align}\frac{d}{dx}f(x)^{2} &= \lim_{h\to 0}\frac{f(x+h)^{2} - f(x)^{2}}{h} \\&= \lim_{h\to 0}\frac{\big(f(x + h) - f(x)\big)\big(f(x + h) + f(x)\big)}{h}\\&=\lim_{h\to 0}\frac{f(x + h) - f(x)}{h}\big(f(x+h) + f(x)\big)\\&=\lim_{h\to 0}\frac{f(x + h) - f(x)}{h}\...


1

As hint:$$\lim_{t \to 0}\frac{f^2(x+t)-f^2(x)}{t}=\\ \lim_{t \to 0}\frac{f^2(x+t)-f(x+t)f(x)+f(x+t)f(x)-f^2(x)}{t}=\\ \lim_{t \to 0}\frac{f(x+t)(f(x+t)-f(x))+f(x)(f(x+t)-f(x))}{t}=\\ \lim_{t \to 0}\frac{f(x+t)(f(x+t)-f(x))}{t}+\lim_{t \to 0}\frac{f(x)(f(x+t)-f(x))}{t}=\\$$can you take over now ?


1

You can use FTOC, and you will obtain $\sin(t^2)$, then just evaluate at your test points.


1

If we simplify $A = \frac {\sqrt 3}{4}(a^2)(2) + (3a)\frac 4{(\sqrt 3)(a^2)},$ we get $A = \frac {\sqrt 3}{2}a^2 + \frac {4\sqrt 3}{a}.$ Now differentiate. Each term is pretty simple, just apply the power rule. $\frac {dA}{da} = \sqrt 3a - \frac {4\sqrt 3}{a^2}$ The $\sqrt 3$ is just a constant, and should be treated like any other constant. It just rides ...


1

No. Define $f$ by $f(x)=0$ for $ x\le0$ and $f(x)=x^2\sin(1/x)$ for $x>0$. One can show $f$ is differentiable and that $f'$ is not continuous at 0. Obviously, the left-hand limit of $f'$ at 0 exists.


1

The first implication is false. $$(\forall x\in \Bbb R)\; \; f(x)-x\le 0$$ implies that the function $ g:x\mapsto f(x)-x $ is negative but not necessarily decreasing $(g'(x)=f'(x)-1\le 0 $ Take for example $$f(x)=x-x^2$$ then $$x>1 \implies f'(x)-1<-2$$ The second is semi true : In fact, For $ x\ge 0$, $$(\forall t\in[0,x])\; f'(t)\le 1\implies$$ $$\...


1

Try $f(x)=\ln (x+1)$. It can be shown that $f(x)\le x$ but $f'(-0.5)=2>1$


1

Of course, it is not always true. As you supposed, let $\dot{x} = f(t,x)$ with $x(0) = x_0$. Then, $$ \dfrac{dT}{dx}(t,x)= \dfrac{\partial T}{\partial t} \dfrac{dt}{dx}+\dfrac{\partial T}{\partial x} $$ which is not guaranteed to be identically zero. Think of a situation where a mass traverses the floor while losing its kinetic energy. Clearly in this case, $...


1

Use the product rule: $\dfrac {d (u v)} {d a} = u \dfrac {dv} {da} + v \dfrac {du} {da}$ In this case, $u = 18^2$ and $v = a$. So you get $18^2 \times \dfrac {da} {da} + a \dfrac {d (18^2)} {da}$. This of course works out to $18^2 \times 1 + a \times 0$.


1

$\frac{1}{x}=x^{-1}$, hence $$\frac{d}{dx}\frac{1}{x}=(-1)x^{-2}=-\frac{1}{x^2}$$


1

If $c$ is a constant, then $(cf(x))'=c(f(x))'$; you get the square because $\frac{1}{x}=x^{-1}$ and $(x^{-1})'=(-1)\cdot x^{-2}=-\frac{1}{x^2}$. To obtain the correct result you have to use these two rules.


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