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Problem (A): Compute the derivative of $f(x)=x^x$.
Solution: Rewrite $x^x$ using $\exp$ and $\ln$ as follows:
$$x^x = e^{\ln(x^x)} = e^{x\ln(x)}.$$
Now differentiate using that formula to get
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} x^x &= \dfrac{\mathrm{d}}{\mathrm{d}x} e^{x\ln(x)} \\
&= e^{x\ln(x)} \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ x \ln(x) \right] \\
&= x^x \left( \ln(x) + x \dfrac{1}{x} \right) \\
&= x^x \left( \ln(x) + 1 \right).
\end{array}$$
Section 5.3 #99: Let $f$ be twice-differentiable and one-to-one on an open interval $I$. Show that its inverse function $g$ satisfies
$$g''(x) = -\dfrac{f''(g(x))}{[f'(g(x))]^3}.$$
Solution: Using Theorem 5.9 (pg. 341), we know that
$$g'(x) = \dfrac{1}{f'(g(x))}.$$
Now calculating another derivative and using the quotient and chain rules gives us
$$g''(x) = \dfrac{f'(g(x))\cdot 0 - (1) f''(g(x))g'(x)}{[f'(g(x))]^2} = -\dfrac{f''(g(x))g'(x)}{[f'(g(x))]^2}.$$
To complete the problem simply plug in what $g'(x)$ is into this formula to get
$$g''(x) = -\dfrac{f''(g(x)) \left( \frac{1}{f'(g(x))} \right)}{[f'(g(x))]^2}=-\dfrac{f''(g(x))}{[f'(g(x))]^3},$$
as was to be shown.
Section 5.4 #133: Find the value of $a$ such that the area bounded by $y=e^{-x}$, the $x$-axis, $x=-a$, and $x=a$ is $\dfrac{8}{3}$.
Solution: The area being described can be calculated using the following integral:
$$\displaystyle\int_{-a}^a e^{-x} \mathrm{d}x = -e^{-x} \Bigg|_{-a}^a = e^{-a}-e^a.$$
We must find the value $a$ with the property that the value of this integral becomes $\dfrac{8}{3}$, in other words, we are being asked to solve the equation
$$e^{-a}-e^a = \dfrac{8}{3}.$$
Simplifying the negative exponent yields
$$\dfrac{1}{e^a} - e^a = \dfrac{8}{3}.$$
Multiply both sides by $e^a$ to transform this into a quadratic equation:
$$1 - e^{2a} = \dfrac{8}{3}e^a.$$
Rearrange the equation for convenience:
$$e^{2a} + \dfrac{8}{3} e^a - 1 = 0.$$
If we let $u=e^a$, then this is simply the quadratic equation $au^2+bu+c=0$ with $a=1$, $b=\dfrac{8}{3}$, and $c=-1$. The easiest way to solve it is to simply use the quadratic formula which yields:
$$\begin{array}{ll}
u &= \dfrac{-\frac{8}{3} \pm \sqrt{(\frac{8}{3})^2-4(1)(-1)}}{2} \\
& =\dfrac{-\frac{8}{3} \pm \sqrt{\frac{64}{9}+4}}{2} \\
&= \dfrac{-\frac{8}{3} \pm \sqrt{\frac{100}{9}}}{2} \\
&= \dfrac{-\frac{8}{3} \pm \frac{10}{3}}{2}
\end{array}$$
Taking the $+$ solution yields $u=\dfrac{\frac{2}{3}}{2}=\dfrac{1}{3}$ and taking the $-$ solution yields $u=\dfrac{-\frac{18}{3}}{2} = -\dfrac{9}{3}=-3$. But $u=e^a$, so now substitute that in to the $+$ solution to get $e^a=\dfrac{1}{3}$, or equivalently $a= \ln \left( \dfrac{1}{3} \right)=-\ln(3)$. Notice that if we try to substitute into the other solution, we run into a problem: we cannot solve the equation $e^a=-3$ because, to do so, we would have to evaluate $\ln(-3)$, which our definition of $\ln$ does not handle (but of course there is a sense in which it works using complex numbers).
Section 5.4 #139: Given $e^x \geq 1$ for $x \geq 0$, it follows that
$$\displaystyle\int_0^x e^t \mathrm{d}t \geq \displaystyle\int_0^x 1 \mathrm{d}t.$$
Perform this integration to derive the inequality
$$e^x \geq 1+x$$
for $x \geq 0$.
Solution: Integrate the left-hand side of the inequality:
$$\displaystyle\int_0^x e^t \mathrm{d}t = e^t \Bigg|_0^x = e^x - e^0 = e^x - 1.$$
Now integrate the right-hand side to get
$$\displaystyle\int_0^x 1 \mathrm{d}t = t \Bigg|_0^x = x-0 = x.$$
This means the given inequality can be rewritten as
$$e^x-1 \geq x,$$
and adding $1$ to both sides yields the desired result
$$e^x \geq x+1,$$
as was to be shown.
