We are given four problems: a) Find $y'$ using implicit differentiation, given $x^y = y^x$. b) A man walks north at 60 cm/s from a point $P$. Five minutes later, a woman walks south at 40 cm/s from a point 1000 cm east of $P$. Find the rate at which the people are moving apart 15 minutes after the woman starts walking. c) Prove that if $\frac{a_n}{n+1} + \frac{a_{n-1}}{n} + \dots + \frac{a_1}{2} + a_0 = 0$, then the equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0$ has at least one real root between 0 and 1. d) Find $f'(x)$ if $\frac{d}{dx} (f(2 \ln x)) = \frac{\ln x}{(\ln x)^2 + 1}$. Simplify the answer.

1. Problem Description

We are given four problems:

a) Find $y'$ using implicit differentiation, given $x^y = y^x$.

b) A man walks north at 60 cm/s from a point $P$. Five minutes later, a woman walks south at 40 cm/s from a point 1000 cm east of $P$. Find the rate at which the people are moving apart 15 minutes after the woman starts walking.

c) Prove that if $\frac{a_n}{n+1} + \frac{a_{n-1}}{n} + \dots + \frac{a_1}{2} + a_0 = 0$, then the equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0$ has at least one real root between 0 and

1. d) Find $f'(x)$ if $\frac{d}{dx} (f(2 \ln x)) = \frac{\ln x}{(\ln x)^2 + 1}$. Simplify the answer.

2. Solution Steps

a)

Given $x^y = y^x$. Taking the natural logarithm of both sides, we have:

$y \ln x = x \ln y$

Now, we differentiate both sides with respect to $x$:

$\frac{d}{dx}(y \ln x) = \frac{d}{dx}(x \ln y)$

Using the product rule:

$\frac{dy}{dx} \ln x + y \cdot \frac{1}{x} = \frac{dx}{dx} \ln y + x \cdot \frac{1}{y} \frac{dy}{dx}$

$\frac{dy}{dx} \ln x + \frac{y}{x} = \ln y + \frac{x}{y} \frac{dy}{dx}$

$\frac{dy}{dx} \ln x - \frac{x}{y} \frac{dy}{dx} = \ln y - \frac{y}{x}$

$\frac{dy}{dx} (\ln x - \frac{x}{y}) = \ln y - \frac{y}{x}$

$\frac{dy}{dx} = \frac{\ln y - \frac{y}{x}}{\ln x - \frac{x}{y}} = \frac{\frac{x \ln y - y}{x}}{\frac{y \ln x - x}{y}} = \frac{y}{x} \cdot \frac{x \ln y - y}{y \ln x - x}$

Since $y \ln x = x \ln y$, we have

$\frac{dy}{dx} = \frac{y}{x} \cdot \frac{x \ln y - y}{x \ln y - x} = \frac{y}{x} \cdot \frac{x \ln y - y}{x(\ln y - 1)}$

If we multiply the numerator and denominator by $x$, we get:

$\frac{y(x \ln y - y)}{x^2(\ln y - 1)}$

$\frac{dy}{dx} = \frac{y(x \ln y - y)}{x(y \ln x - x)} = \frac{y(x \ln y - y)}{x(x \ln y - x)}$

b)

Let $y_m(t)$ be the distance the man has walked north from point $P$ at time $t$.

Let $y_w(t)$ be the distance the woman has walked south from the point 1000 cm east of $P$ at time $t$.

The man starts walking at $t=0$ and the woman starts walking at $t=300$ seconds.

$y_m(t) = 60t$

$y_w(t) = 40(t - 300)$ for $t \ge 300$

Let $d(t)$ be the distance between the man and the woman. We can use the Pythagorean theorem to find $d(t)$. The horizontal distance between them is 1000 cm. The vertical distance between them is $y_m(t) + y_w(t)$.

$d(t) = \sqrt{1000^2 + (y_m(t) + y_w(t))^2}$

$d(t) = \sqrt{1000^2 + (60t + 40(t - 300))^2}$

$d(t) = \sqrt{1000^2 + (100t - 12000)^2}$

We want to find $\frac{dd}{dt}$ when $t = 15 \times 60 = 900$ seconds after the woman starts walking, i.e., when $t=300+900 = 1200$ seconds.

$\frac{dd}{dt} = \frac{1}{2} (1000^2 + (100t - 12000)^2)^{-1/2} \cdot 2(100t - 12000) \cdot 100$

$\frac{dd}{dt} = \frac{100(100t - 12000)}{\sqrt{1000^2 + (100t - 12000)^2}}$

At $t = 1200$:

$\frac{dd}{dt} = \frac{100(100 \cdot 1200 - 12000)}{\sqrt{1000^2 + (100 \cdot 1200 - 12000)^2}}$

$\frac{dd}{dt} = \frac{100(120000 - 12000)}{\sqrt{1000^2 + (120000 - 12000)^2}}$

$\frac{dd}{dt} = \frac{100(108000)}{\sqrt{1000000 + (108000)^2}}$

$\frac{dd}{dt} = \frac{10800000}{\sqrt{1000000 + 11664000000}} = \frac{10800000}{\sqrt{11665000000}} = \frac{10800000}{108009.258} \approx 99.99$ cm/s

$\frac{dd}{dt} \approx 100$ cm/s

c)

Let $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$.

Consider the function $Q(x) = \frac{a_n}{n+1} x^{n+1} + \frac{a_{n-1}}{n} x^n + \dots + \frac{a_1}{2} x^2 + a_0 x$.

Then $Q'(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = P(x)$.

We are given that $Q(1) = \frac{a_n}{n+1} + \frac{a_{n-1}}{n} + \dots + \frac{a_1}{2} + a_0 = 0$. Also $Q(0) = 0$.

Since $Q(0) = Q(1) = 0$, by Rolle's theorem, there exists a $c \in (0, 1)$ such that $Q'(c) = 0$.

But $Q'(x) = P(x)$, so $P(c) = 0$ for some $c \in (0, 1)$. Therefore, $P(x)$ has at least one real root between 0 and

1.

d)

We are given $\frac{d}{dx} (f(2 \ln x)) = \frac{\ln x}{(\ln x)^2 + 1}$.

Let $u = 2 \ln x$, so $\ln x = \frac{u}{2}$. Then $\frac{d}{dx} (f(u)) = \frac{u/2}{(u/2)^2 + 1} = \frac{u/2}{\frac{u^2}{4} + 1} = \frac{2u}{u^2 + 4}$.

By the chain rule, $\frac{d}{dx} (f(2 \ln x)) = f'(2 \ln x) \cdot \frac{d}{dx} (2 \ln x) = f'(2 \ln x) \cdot \frac{2}{x}$.

Thus, $f'(2 \ln x) \cdot \frac{2}{x} = \frac{\ln x}{(\ln x)^2 + 1}$.

$f'(u) \cdot \frac{2}{x} = \frac{u/2}{(u/2)^2 + 1}$

$x = e^{u/2}$, so $f'(u) \cdot \frac{2}{e^{u/2}} = \frac{u/2}{u^2/4 + 1} = \frac{2u}{u^2 + 4}$

$f'(u) = \frac{2u}{u^2 + 4} \cdot \frac{e^{u/2}}{2} = \frac{u e^{u/2}}{u^2 + 4}$

Therefore, $f'(x) = \frac{x e^{x/2}}{x^2 + 4}$.

3. Final Answer

a) $\frac{dy}{dx} = \frac{y}{x} \cdot \frac{x \ln y - y}{y \ln x - x}$

b) 100 cm/s

c) See solution above.

d) $f'(x) = \frac{x e^{x/2}}{x^2 + 4}$