5
Hint: Let $I=[-1,1]$. $f(t)=t^2$, $g(t)=t\left|t\right|$.
4
No, the given conditions do not imply that $f(e^x)$ is convex in some interval $(-\delta, 0]$. We will give a concrete counterexample below.
Preliminaries/motivation
If $f$ is differentiable then $g(u) = f(e^u)$ is convex iff $h(x) = xf'(x)$ is increasing. We have that $h$ is negative and $h(1) = 0$. Of course that does not imply that $h$ is increasing near $...
4
Since the partial derivatives exist and are continuous at every point, then the function is differentiable at every point.
4
Hint: Rewrite the DE as $$\frac{dx}{dy}-\frac{x}{y}=y$$ which is easy by integrating factor method
4
$$\frac{du}{u}=-\ln(u(t))dt$$
You cannot integrate $\int-\ln(u(t))dt$ directly.
The mistake is : You integrate $\quad \int-\ln(u(t))du= -u\ln(u)+u\quad$ but this is not $\quad \int -\ln(u(t))dt$.
Back to the original ODE :
$$\frac{du}{dt}=u\ln(u)$$
This is a separable ODE thus
$$\frac{du}{u\ln(u)}=dt$$
$$\int\frac{du}{u\ln(u)}=\int dt$$
$$\ln(\ln(u))=t+c$$
$$...
3
Counterexample. Let $f(x) = \frac 1x \sin(x^2)$. Then $f(x) \to 0$ as $x \to \infty$, but
$$
f'(x) = - \frac{1}{x^2} \sin(x^2) + \frac 1x \cos(x^2) \cdot 2x
= - \frac{1}{x^2} \sin(x^2) + 2 \cos(x^2),
$$
which is not convergent to $0$ as $x \to \infty$. The problem here lies of course in the second derivative, which is not bounded.
Problem with your ...
3
The instantaneous speed is $s(t)=\sqrt{x'(t)^2+y'(t)^2}$, and the average speed is $\frac{1}{T}\int_0^Ts(t)\,\mathrm d t$, where $T=20 \,\text{s}$.
Does it make sense?
3
$\dfrac{x^2-1}{(x-1)(x-2)(x-3)}= \dfrac{x+1}{(x-2)(x-3)} = \dfrac{(x-2)+3}{(x-2)(x-3)}= \dfrac{1}{x-3}+\dfrac{3}{(x-2)(x-3)}= \dfrac{1}{x-3}+3\left(\dfrac{1}{x-3}-\dfrac{1}{x-2}\right)= \dfrac{4}{x-3}-\dfrac{3}{x-2}$. Can you take it from here?
3
Besides from the solution proposed by @SonGohan, you can prove it by the definition.
The candidate to be the derivative is given by $L(x,y) = (3x^{2} - 3y^{2},-6xy)$.
Now you can prove that
\begin{align*}
\lim_{(x,y)\to(0,0)}\frac{|f(x,y) - f(0,0) - L(x,y)((x,y) - (0,0))^{T}|}{\|(x,y)\|} = 0
\end{align*}
Indeed, this is the case.
To begin with, notice the ...
2
You are right: since $u_x(0,0)=u_y(0,0)=v_x(0,0)=v_y(0,0)=0$, $(0,0)$ is a solution of the Cauchy-Riemann equations (and, since the partial derivatives are continuous at $(0,0)$, $f$ differentiable at $0$).
2
$\frac {d\cos^4 u}{du}\ne -\sin^4 u$.
Let $v(x) = x^4$ then we have
SO if you have $\cos^4 u = v(\cos u)$ then
$\frac {d\cos^4 u}{du} = \frac {d v(\cos u)}{du} = \frac {d v(\cos u)}{d(\cos u)} \frac d{\cos u}{du}=$
$4(\cos u)^3\cdot (-\sin u)=$
$-4\sin u\cos^3 u$.
Doing it via the chain rule is
$[\frac{1}{7}cos^4(x^2+2x+3)]'=$
$\frac 17[4(\cos^3(x^2+2x+3))(-\...
2
Note that each curve represents a set of three equally spaced pedals with each pedal subtending a polar angle of $\frac\pi3$.
Thus, the area is
$$\frac{3}{2}\int_0^{\frac\pi3} (f(\theta)^2-g(\theta)^2)d\theta
= \frac{3}{2}\int_0^{\frac\pi3} (64\sin^2(3\theta)-25\sin^2(3\theta))d\theta=\frac{39\pi}4
$$
2
It seems a lit bit hard but if you try to find the pattern from n=1 to n=2 ,
so $(e^x (x^2 + x))\prime$ = $e^x (x^2 + 3x +1)$ this is for n egale to 1 you cannot get the pattern , so for $n = 2\quad , (e^x(x^2 + x))\prime\prime = e^x (x^2 + 5x + 4 )$ , to deduce calculate for $n=3$ :
$(e^x(x^2+x))\prime\prime\prime$ = $e^x (x^2 + 7x + 9 )$ , the patern is $\...
2
Just by calculating $f',f''$ and $f'''$ by hand, you can see the following pattern:
$$\frac{\mathrm{d}^nf}{\mathrm{d}x^n}(x) = e^x(x^2+(2n+1)x+n^2).$$
Now, we can prove this by induction (the base steps are already done if you did the calculations). Assume $\frac{\mathrm{d}^nf}{\mathrm{d}x}(x) = e^x(x^2+(2n+1)x+n^2)$ for some $n\geq 1$, then $$\frac{\mathrm{...
2
$m$ can be found without using slope. The given line touches the curve at $(x,2x-4)$ so that
$$2x-4=(m+3)x^2+mx$$
is satisfied. This is a quadratic in $x$,
$$(m+3)x^2+(m-2)x+4=0$$
and its discriminant is zero, since for a given $m$, the quadratic has a unique root (tangency point). Thus
$$D : (m-2)^2-4\cdot 4 \cdot (m+3)=0$$
$$\Rightarrow m^2-20m-44=0 \...
2
If you want to figure out where $g'$ is zero, compute it as
\begin{align*}
g'(x) &= -3\int_a^b\, (f(t) - x)^2\, dt \\
&= -3\int_a^b\, f^2(t)\, dt + 6x \int_a^b f(t)\, dt -3x^2(b - a).
\end{align*}
It's a quadratic in $x$. Set it to zero and solve for the $x$s that are the critical points.
If you set $I_0 = b - a$, $I_1 = \int_a^b\, f(t)\, dt$, $I_2 = ...
2
$$\sum^{\infty}_{n=1} (-1)^n\frac{t^{(2n+1)}}{(2n+1)!}= \sin (t)-t$$
So, as you noticed, the problem is "simply" to compute
$$I=\int \sin(\sin(x))\,dx$$ which cannot be obtained aven using special functions.
If it was for definite integrals, the problem would be different since
$$\int_0^{\frac \pi 2} \sin(\sin(x))\,dx=\frac{\pi }{2}\pmb{H}_0(1)$$ ...
2
The graph of $f(x)=x^2$ has a minimum at $x=0$.
The graph of $f(x)=-x^2$ has a maximum at $x=0$.
The graph of $f(x)=x\sin x$ does not have a minimum or maximum at $x=0$. Rather, it has a local minimum.
The graph of $f(x)=x^3$ does not have a minimum, maximum, local minimum, or local maximum at $x=0$. It has a point of inflection.
Just knowing that $f'(x)=0$ ...
1
A function which is differentiable is continuous, $f_n$ is differentiable (continuous) since it is a composition of differentiable (continuous) functions $g_n(x)=x^2+{1\over n^2}$ and $h(x)=\sqrt{x}$ (defined on $(0,+\infty)$.
1
Yes, it is valid to prove that a function is continuous by actually proving that it is differentiable.
However, it seems more natural for me to see that$$f_n'(x)=\frac x{\sqrt{x^2+1/n^2}}$$and to say that this function is continuous since it is the quotient of two continuous functions.
1
Your work for this part is correct so far. The derivative is $g'(x) = 3x^2 - 10x$. Recall that the equation $y = 13x$ has the slope-intercept form, which is a straight line with slope $13$.
We solve for the value of $x$ such that $g'(x) = 13$. We have
\begin{align}
3x^2 - 10x &= 13 \\
3x^2 - 10x - 13 &= 0 \\
x &= -1 \text{ or } \frac{13}{3}.
\end{...
1
What you did is correct.
You could improve readability by stating that for a convex map the slope of the line joining a fixed point with abscissa $a$ to a point with abscissa $x \gt a$ increases with $x$. Applying that in your case with $a=0$ and knowing that
$$\frac{f(x) - f(0)}{x- 0} = \frac{f(x)}{x}$$ you get that $g$ is increasing.
The introduction of $b$...
1
Let $y=f(x)$ and let $x(t)$ be a given function. Notice that $\frac{dy}{dx}=f'(x)$, so we need to calculate $\frac{d}{dt}(f'(x(t)).$ Applying the chain rule, we find this is $f''(x(t))\cdot \frac{dx}{dt}$. Translating back to your notation, this means that $\frac{d}{dt} \frac{dy}{dx}=\frac{d^2 y}{dx^2} \frac{dx}{dt}.$
1
Let $\Pi_{j=0}^i K_j=a_i$ for $i=1\dots p$, then
$$f'(x,k)=x^{p-k-1}a_k\frac{(p-k)\sum_{i=1}^p x^{p-i}a_i-x\sum_{i=1}^p (p-i)x^{p-i-1}a_i}{\left(\sum_{i=1}^p x^{p-i}a_i\right)^2}
$$
Now, $$f'(x,k)=0\implies(p-k)\sum_{i=1}^p x^{p-i}a_i-x\sum_{i=1}^p (p-i)x^{p-i-1}a_i=0,$$
and, therefore
$$
\begin{aligned}(p-k)\sum_{i=1}^p x^{p-i}a_i&=x\sum_{i=1}^p (p-i)x^{...
1
The problem is poorly phrased. It awkwardly mixes real and complex differentiability as well as pointwise and local notions. Let me try to clear things up.
As a map $\mathbb R^2\to\mathbb R^2$ the map $f(z)=z^2\bar z$ has polynomials as component functions. In particular, it is real differentiable everywhere and even real analytic (meaning that all ...
1
$\lim_{x\rightarrow \infty}\frac{y^2}{x^2}=3$ is a result you derived from the original equation. It contains relationship between x and y. We know that x is a set of constant values in R. In order for the equation to be true, y has to be the same order with x.
For example, if $y=x^2$ then $\lim_{x\rightarrow \infty}\frac{y^2}{x^2}=\lim_{x\rightarrow \infty}\...
1
I will write $D_x$ instead of $\frac{\mathrm{d}}{\mathrm{d}x}$.So, you want to calculate
$$D_x\left[ \frac{1}{7}\cos^4(x^2+2x+3)\right]=\frac{1}{7}D_x\left[\cos^4(x^2+2x+3)\right].$$
I will leave the $1/7$ and multiply it again at the end. Set $g=\cos(x^2+2x+3)$. Now that's
$$D_g\left[g^4\right]\cdot D_x \left[\cos(x^2+2x+3)\right] = 4g^3 D_x \left[\cos(x^2+...
1
The function that you hace to derive is the composition of three two function: $f$, $g$, $h$.
In effect $\frac{1}{7}\cos^4(x^2+2x+3)=\frac{1}{7}h(g(f(x)))$, with $f(x)=x^2+2x+3$, $g(x)=cos(x)$, $h(x)=x^4$.
So applying the chain rule of derivation to this composition the derivative will be =$-\frac{4}{7}\cos^3{(x^2+2x+3)}\sin{(x^2+2x+3)}*(2x+2)$
1
$\def\T{\operatorname{Tr}}\def\p#1#2{\frac{\partial #1}{\partial #2}}$For
typing convenience, let's define the matrix variable
$$B=S\circ S\circ A$$
and use a colon to denote the trace/Frobenius product
$$\eqalign{
A:B &= \T(A^TB) \\
A:A &= \big\|A\big\|_F^2 \\
}$$
The properties of the trace allow the terms in such a product to be rearranged
$$\...
1
Let $C = \lim\limits_{x \to 0} f(x)$. Obviously for any function $h$, we have
$$\lim\limits_{x \to 0^+} h(x) = \lim\limits_{x \to 0^-} h(-x).$$
Then
$$C = \lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^-} f(-x) = \lim\limits_{x \to 0^-} -f(x) = - \lim\limits_{x \to 0^-} f(x) = -C $$
so $C = 0$.
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