The lecture attendances are independent Bernoulli trials with success probability $f$. Their count has the binomial distribution:

By symmetry, each student in the university has probability $(S/N)(1/S)=1/N$ of being in the college and then selected. Hence the prior probability is

The domain is $x>0$, and $x\log x\to 0$ from below as $x\downarrow 0$. Its derivative is $1+\log x$, so it decreases up to $x=e^{-1}$ and then increases, with its unique global minimum at $(e^{-1},-e^{-1})$. It crosses the horizontal axis at $(1,0)$ and tends to $+\infty$ as $x\to \infty$. Its second derivative is $1/x>0$, so it is strictly convex. The right panel of the preceding figure shows these features; zero is an excluded endpoint with a finite limiting value, rather than a vertical asymptote.

Write $x/(x-1)=1+1/(x-1)$. The graph is a translated hyperbola, with vertical asymptote $x=1$ and horizontal asymptote $y=1$. It passes through $(0,0)$, lies below $y=1$ for $x<1$ and above it for $x>1$, and has derivative $-1/(x-1{)}^2<0$ on each component of its domain. The left panel below shows both branches.

Integration by parts with $u=x$ and $dv=e^{x}dx$ gives

Use the change of variables $u=x^2+1$, $du=2xdx$. ThenDifferentiating verifies this antiderivative on the whole real line.

For real square-integrable $f,g$, nonnegativity of $\int (f-\lambda g{)}^{2}dx$ gives a quadratic in the real number $\lambda$. If $\int g^{2}dx>0$, choose $\lambda =\int fgdx/\int g^{2}dx$ to obtainIf $\int g^{2}dx=0$, then $g=0$ almost everywhere and the same inequality is immediate. This proves the integral Cauchy-Schwarz inequality.

Put $T=t_f-t_i>0$ and $L=x_f-x_i>0$. The forward speed satisfies $L={\int }_{t_i}^{t_f}vdt$, so $V_t=L/T$. The change of variables $dx=vdt$ gives $V_x=L^{-1}{\int }_{t_i}^{t_f}{v}^{2}dt$. Apply the Cauchy-Schwarz inequality to $v$ and the constant function one:Equality holds exactly when $v$ is constant almost everywhere. Equivalently, the time and distance averages of speed satisfy $V_x-V_t={\mathrm{Var}}_t(v)/V_t$, which identifies the excess as a measure of speed variation.

For functions with the required derivatives, the Leibniz rule isAt $n=0$ it is immediate. If it holds at $n$, differentiating each summand gives two sums. The coefficient of $f^{(k)}{g}^{(n+1-k)}$ after regrouping is $\left(\genfrac{}{}{0ex}{}{n}{k}\right)+\left(\genfrac{}{}{0ex}{}{n}{k-1}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)$ by Pascal's identity, including the two endpoint terms. This proves the formula by mathematical induction.

Apply it to $(fg)h$ and then expand the derivatives of $fg$. For $p+q+r=n$, the resulting coefficient is $\left(\genfrac{}{}{0ex}{}{n}{p+q}\right)\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)=n!/(p!q!r!)$. Consequentlywhere $p,q,r$ range over nonnegative integers. This is the three-factor generalized Leibniz rule.

The quotient rule and the derivative of the exponential function give

The chain rule gives

If $A>B$, divide numerator and denominator by $e^{A/x^2}$. The expression becomes $(A+B{e}^{-(A-B)/x^2})/(1+e^{-(A-B)/x^2})$, which tends to $A$ since $A-B>0$. If $B>A$, the same argument gives $B$. If $A=B$, the expression equals $A$ for every $x\ne 0$. ThusThis is the two-term case of exponential concentration of a weighted mean, and holds for negative as well as positive parameters.