Daily math problems for July 27, 2026 — Calculus, Analytic Geometry, Algebra & more

$\bullet$ Differentiate by the chain rule:

$$f^{\mathrm{\prime }}(x)=6\pi \cos (2\pi x+\frac{\pi }{4})$$

$\bullet$ Differentiate again:

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=-12{\pi }^2\sin (2\pi x+\frac{\pi }{4})$$

$\bullet$ Evaluate at $x=\pi \mathrm{/}3$ and $x=\pi \mathrm{/}6$:

$$f^{\mathrm{\prime }}\text{\hspace{0.17em}⁣}\left(\frac{\pi }{3}\right)=6\pi \cos \text{\hspace{0.17em}⁣}(\frac{2{\pi }^2}{3}+\frac{\pi }{4})$$

$$f^{\mathrm{\prime }\mathrm{\prime }}\text{\hspace{0.17em}⁣}\left(\frac{\pi }{6}\right)=-12{\pi }^2\sin \text{\hspace{0.17em}⁣}(\frac{{\pi }^2}{3}+\frac{\pi }{4})$$

$\bullet$ Add the two:

$$6\pi \cos \text{\hspace{0.17em}⁣}(\frac{2{\pi }^2}{3}+\frac{\pi }{4})-12{\pi }^2\sin \text{\hspace{0.17em}⁣}(\frac{{\pi }^2}{3}+\frac{\pi }{4})$$

$m_1\cdot m_2=-1\Rightarrow m_2=-{\displaystyle \frac{1}{2}}$.

A $3$-cycle has sign $(-1{)}^{3-1}=+1$. (Equivalently, it can be written as the product of two transpositions: $(123)=(13)(12)$.)

${\mathbb{Z}}_n$ is a field iff every non-zero element is invertible, iff every element from $1$ to $n-1$ is coprime to $n$, iff $n$ is prime.

By parts: $\int x\ln xdx={\displaystyle \frac{x^2}{2}}\ln x-\int {\displaystyle \frac{x}{2}}dx={\displaystyle \frac{x^2}{2}}\ln x-{\displaystyle \frac{x^2}{4}}$. Evaluating on $[0,1]$ (with $x^2\ln x\to 0$ as $x\to 0^{+}$) gives $0-\frac{1}{4}-0=-\frac{1}{4}$.

The exponent ${\displaystyle \frac{6-3k}{2}}$ is an integer when $k$ is even: $k\in \{0,2,4,6\}$ — four values.

${\displaystyle \frac{2x^2+3x+1}{x-1}}=2x+5+{\displaystyle \frac{6}{x-1}}$. As $x\to \pm \mathrm{\infty }$, the remainder vanishes, leaving the slant asymptote $y=2x+5$.

The positive divisors of $12$ are $\{1,2,3,4,6,12\}$, so $\mathrm{\mid }A\mathrm{\mid }=6$.

$\overrightarrow{AB}=(4,0)$ and $\overrightarrow{DC}=(4,0)$ — same vector, so $ABCD$ is a parallelogram. Base $=4$, height $=3$. Area $=12$.

$\bullet$ Take reciprocals of each equation, using $\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$:

$$\frac{1}{x}+\frac{1}{y}=\frac{3}{5},\frac{1}{x}+\frac{1}{z}=\frac{2}{3},\frac{1}{y}+\frac{1}{z}=\frac{1}{4}$$

$\bullet$ Let $a=1\mathrm{/}x$, $b=1\mathrm{/}y$, $c=1\mathrm{/}z$. Sum all three:

$$2(a+b+c)=\frac{3}{5}+\frac{2}{3}+\frac{1}{4}=\frac{91}{60}$$

$$a+b+c=\frac{91}{120}$$

$\bullet$ Subtract each pair-sum from $a+b+c$:

$$a=\frac{61}{120},b=\frac{11}{120},c=\frac{19}{120}$$

$\bullet$ Recover $x,y,z$ and add:

$$x+y+z=\frac{120}{61}+\frac{120}{11}+\frac{120}{19}$$