7

Hint : Show inductively that $$f'(x)=\left[\cos(x)\right]\times\left[\cos(\sin(x))\right]\times\left[\cos(\sin(\sin(x))\right]\times\left[...\right]\times\left[\cos(\sin(\sin(\sin(...(x)))))\right]$$


6

A possible way is to set $x = \tan t$ with $t \in \left(0,\frac{\pi}{2}\right)$. So, the inequality is equivalent to $$\frac{\tan t}{1+\frac 2{\pi} \tan t}< t \text{ for } t \in \left(0,\frac{\pi}{2}\right)$$ or after rearranging $$ \tan t < \frac t{1-\frac 2{\pi}t} \text{ for } t \in \left(0,\frac{\pi}{2}\right)$$ But this is true because for $t \in ...


6

Discussion of L'Hopital's Rule aside, the straightforward way to solve the limit would be to multiply both numerator and denominator by $\frac1x$: $$L = \lim_{x \to \infty} \frac{x}{\sqrt{(x+a)(x+b)}} = \lim_{x \to \infty} \frac{1}{\sqrt{\frac {(x+a)(x+b)}{x^2}}}=\lim_{x \to \infty} \frac{1}{\sqrt{\left(1+\frac ax\right)\left(1+\frac bx\right)}}=1$$


6

All the other answers are for the case where there are infinite x's. However, there are finite ($n$) x's in this problem! Let's first denote the function as $f_n(x)=x^{x^{\cdots^x}}$ with $n$ x's in the exponent. For example, $f_1(x)=x^x, f_0(x)=x.$ Take the logarithm of both sides, we get $\ln f_n(x)=f_{n-1}(x)\ln x$. Taking the derivative gives $$\frac{f'...


6

$$\int_0^1 xf(x)\,dx=\int_0^1\left(x-\frac12\right)f(x)\,dx=\underbrace{-\frac12 x(1-x)f(x)\Bigg|_0^1}_{=0}+\frac12\int_0^1 x(1-x)f'(x)\,dx.$$ Hence, if $\displaystyle M=\sup_{x\in(0,1)}|f'(x)|$, then $\displaystyle\left|\int_0^1 xf(x)\,dx\right|\leqslant\frac{M}{2}\int_0^1 x(1-x)\,dx=\frac{M}{12}$.


5

Intuitively, what you want here is a function that iterates between (at least) two values and also gets straighter as time passes. You can think of the usual sine function as a spring, and imagine stretching it to the right. Try to understand why this make the function get straight as $x$ increases. Consequently, $\sin(\sqrt x)$ would work. I suggest you ...


5

Hint : $$f = g \circ h$$ where $$g(x)=\int_{-2020}^{x} \sin \left(\pi t^{2}\right)dt \quad \text{and} \quad h(x)=x^3$$


4

Formally, you should say : second order derivative of y w.r.t. x


4

First, consider $$ f(x)=\frac{\sin(x^2)}{x}. $$ As $x\rightarrow\infty$, this approaches $0$. On the other hand, $$ f'(x)=\frac{2x^2\cos(x^2)-\sin(x^2)}{x^2}=2\cos(x^2)-\frac{\sin(x^2)}{x^2} $$ does not have a limit as $x\rightarrow\infty$. Therefore, the condition on $f'''$ is necessary for the result. The problem with your setup is that you never specify ...


4

Let $f(x)=-x^2$ when $x\le 0$ and $f(x)=0$, when $x>0$.


4

Let us write \begin{equation}g (x) = x h (x)\end{equation} so that $h (x) \rightarrow 1$ when $x \rightarrow 0$. Let us first substitute $s = t u$ in the integral \begin{equation}I \left(x\right) \colon = \int_{{x}^{3}}^{{x}^{2}}\left(\int_{0}^{t}g \left({s}^{2}\right) d s\right) d t = \int_{{x}^{3}}^{{x}^{2}}\left(\int_{0}^{t}{s}^{2} h \left({s}^{2}\...


4

This is an answer based on my comments. Let $$G(t) =\int_{0}^{t}g(s^2)\,ds,F(u)=\int_{0}^{u}G(t)\,dt$$ Then we have $$F'(u) =G(u), G'(t) =g(t^2)$$ and hence by two applications of LHospital Rule we get $$\lim_{u\to 0}\frac{F(u)}{u^4}=\lim_{u\to 0}\frac{G(u)}{4u^3}=\lim_{u\to 0}\frac{g(u^2)}{12u^2}=\frac {1}{12}$$ The limit in question is $$\lim_{x\to 0}\frac{...


4

Hint: if $f'(x) \neq 0$ on $[a,b),$ then $f'(x)>0$ or $f'(x)<0$ on $[a,b).$ What can you conclude from this information? (We don't implicitly require $f'$ to be continuous here! See theorem 5.12 in Rudin's Principles.) Edit: Suppose that $f'(x)>0$ on $[a,b).$ Then by the mean value theorem, $f$ is increasing. It follows that $\lim_{x \to b^-} f(x) = ...


4

Just to flesh out @TheSilverDoe's answer: Denoting functional composition by $\circ$, we want to differentiate $\sin^{\circ n}x$. We prove by induction$$\frac{d}{dx}\sin^{\circ n}x=\prod_{j=0}^{n-1}\cos(\sin^{\circ j}x).$$For $n=0$, this is the trivial $\frac{d}{dx}x=1$, the RHS being an empty product. If the case $n=k$ works, by the chain rule$$\begin{align}...


4

Let $ax^{n}$ be the leading coefficient, then the leading coefficient of both sides of the equation are equal giving: $a(anx^{n-1})^{n} = an(ax^{n})^{n-1}$ So $a^{n+1}n^{n} = a^{n}n$ So $a= \frac{1}{n^{n-1}}$ So one such family is $\frac{x^{n}}{n^{n-1}}$ Now consider the constant coefficient K and the linear coefficient L: $P'(K) = P(L)$ So if $P(x) = Lx+K$ ...


3

Three ways of reading $\dfrac {d^2 y}{dx^2}$... "dee squared y over dee x squared" (your way) "dee two y over dee x two" (informal) The second derivative of $y$ with respect to $x$ (formal)


3

For $1 < a < x < y$ there are $c_1 \in(a, c)$ and $c_2 \in (x, y)$ such that $$ \frac{f(y)-f(x)}{y-x} = f'(c_2) \le f'(c_1) = \frac{f(x)-f(a)}{x-a} \le \frac{f(x)}{x-a} \, . $$ This inequality can be rearranged to $$ \frac{f(x)}{x-a} \ge \frac{f(y)}{y-a} $$ which shows that $\frac{f(x)}{x-a}$ is decreasing. It is also bounded below by zero, so that ...


3

Here's a proof that shows how the condition on the 3rd derivative could be utilized. Take Taylor's remainder theorem to 3rd order and apply it to the intervals $(x,x+h)$, $(x,x+k)$, $h\neq k$. Now consider only the first interval since the second one can be done similarly. There exists $\xi\in(x,x+h)$ such that $$f(x+h)=f(x)+f'(x)h+\frac{f''(x)}{2!}h^2+\frac{...


3

You applied the product rule incorrectly. You should have $$ (uv)'=u'v+uv' $$ where $u=r^2$ and $v=d\theta/dt$. Note that $$ u'=2r r'\quad v'=\frac{d^2\theta}{dt^2} $$ where $\displaystyle(\cdot)'=\frac{d(\cdot)}{dt}$. Notes. \begin{align} \frac{1}{2r}\frac{d}{dt}\left(r^{2}\left(\frac{d\theta}{dt}\right)\right) &=\frac{1}{2r}(2r\frac{dr}{dt}\frac{d\...


3

Let $f_1(x)=x$ and $f_n(x)=x^{f_{n-1}(x)}=x^{x^{\ldots^{x}}}$ ($n$ $x's$ on the exponent). Then $\ln(f_n(x))=f_{n-1}(x)\cdot\ln(x)$ so: $$\frac{f'_n(x)}{f_{n}(x)}=f'_{n-1}(x)\cdot\ln(x)+\frac{f_{n-1}(x)}{x}\implies f'_{n}(x)=f_{n}(x)\cdot\left(f'_{n-1}(x)\cdot\ln(x)+\frac{f_{n-1}(x)}{x}\right).$$


3

Use a substitution $x\mapsto r/x$. It may be easier to follow if you write $y=r/x$ so the right-hand side transforms to $\int_0^\infty e^{-r^2/y^2-y^2}dy$.


3

The derivative of $\int_0^xf(t)dt$ is $f(x)$. In this case, $f(x,y)=\int_0^x{t^2}dt-\int_0^ye^{t^2}dt$ We deduce that $f_x=e^{x^2}$ and $f_y=-e^{y^2}$.


3

The total of their perimeters is $4x + 2\pi r = 2 \implies r = \frac{1-2x}{\pi}$ Total area $A = x^2 + \pi r^2 = x^2 + \frac{(1-2x)^2}{\pi}$ $ = \frac{4+\pi}{\pi} [x^2 - \frac{4}{4+\pi}x + \frac{1}{4+\pi}]$ $ = \frac{4+\pi}{\pi} [(x - \frac{2}{4+\pi})^2 + \frac{\pi}{(4+\pi)^2}]$ That clearly shows that the total area is minimum when $x = \frac{2}{4+\pi}$, so ...


2

Suppose that $f:\Bbb R\to\Bbb R$ and $f+f''=0$. Then $g=f^2+(f')^2$ is constant. Proof: $g'=2ff'+2f''f'=2f'(f+f'')=0$. It's amusing to note that this proves uniqueness for the solution: Cor. If $f(0)=f'(0)=0$ then $f=0$. And hence Cor. If $f''+f=0$, $f(0)=a$ and $f'(0)=b$ then $f(x)=a\cos(x)+b\sin(x)$. Proof: The previous corollary shows that $g=0$, if ...


2

Because this is NOT a "constant function"! The fact that $f(0)= 0$ does not make it a "constant function". Every function has some specific value at $x= 0$ (or at any $x$). The fact that the value happens to be $0$ does not matter.


2

Basically you get the integral $$\int_0^3 \sqrt{1+\left(\frac{\ln(x+2)}{x^2+1}\right)^2}\mathrm{d}x$$ Numerical integration yields $\approx 3.3197.$


2

Yes, you have applied the arc length formula correctly: $$ \int_a^b\sqrt{1+(f'(x))^2}\,dx $$ You then got an elliptic integral and you have the correct approximate value.


2

Follow what you left off, $2x^2 + (x^2-1)\ln(x^2-1)= 2(x^2-1)+(x^2-1)\ln(x^2-1)+2= 2t+t\ln t + 2=h(t), t = x^2-1>0$. Computing $h'(t) = 3+\ln t= 0 \iff t=e^{-3}$. Observe that $h''(t) = \dfrac{1}{t} > 0, t > 0$. Thus $h_{\text{min}} = h(e^{-3})= 2e^{-3}+2+-3e^{-3}= 2-e^{-3} > 0$. By calculus's $2$nd second derivative test, it shows that $h(t) >...


2

Informally, when we say that the 'derivative of sine is cosine', what we mean is that the following quotient is equal to $\cos x$: $$ \frac{\sin(x+dx)-\sin(x)}{dx} $$ where $dx$ is an infinitely small non-zero number. In other words, the line joining the two points $(x,\sin x)$ and $(x+dx,\sin(x+dx))$ has a slope of $\cos x$. Symbolically, if $f(x)=\sin x$, ...


2

The idea is that either $f(x) = 0$ (in which case the inequality holds trivially) or $|f(x)| = f(x)$ or $|f(x)| = -f(x)$ in a neighbourhood of $x$ (in which case we can compute the second derivative). Consider first the case that $f(x_0) > 0$. Then $|f(x)|^3 = f(x)^3 > 0$ for all $x$ in an open interval $I$ containing $x_0$. It follows that $$ 0 \le \...


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