Derivatives of Exponential and Logarithmic Functions

(a) $u=-2x$, $u'=-2$, so $y'=e^{u}u'=-2e^{-2x}$. An exponential is defined for every real $x$.

(b) $u=3x+4$, $u'=3$, so $y'=\dfrac{u'}{u}=\dfrac{3}{3x+4}$. The logarithm needs $3x+4>0$, i.e. $x>-\frac43$.

(c) Change of base: $\log_{5}x=\dfrac{\ln x}{\ln 5}$, and $\dfrac{1}{\ln 5}$ is a constant, so $y'=\dfrac{1}{x\ln 5}$ on $x>0$. The $\ln 5$ does not cancel.

Domain: $e^{x}$ and $2^{x}$ are defined for every real $x$, but $\ln x$ needs $x>0$, so the domain is $(0,\infty)$.

Differentiate term by term with $\frac{d}{dx}e^{x}=e^{x}$, $\frac{d}{dx}\ln x=\frac1x$ and $\frac{d}{dx}a^{x}=a^{x}\ln a$:

$f'(x)=5e^{x}-\frac{3}{x}+2^{x}\ln 2$

At $x=1$: $f'(1)=5e-3+2\ln 2\approx 13.5914-3+1.3863=11.9777$. The $\ln 2$ belongs to $2^{x}$ alone — it never attaches to $e^{x}$.

Product rule: $y'=e^{x}\ln x+e^{x}\cdot\frac1x=e^{x}\left(\ln x+\frac1x\right)$.

At $x=1$: $y(1)=e\cdot 0=0$ and $y'(1)=e(0+1)=e$, so through $(1,0)$ with slope $e$,

$y=e(x-1)$

Product rule again: $y''=e^{x}\left(\ln x+\frac1x\right)+e^{x}\left(\frac1x-\frac{1}{x^{2}}\right)=e^{x}\left(\ln x+\frac2x-\frac{1}{x^{2}}\right)$, so $y''(1)=e(0+2-1)=e$.

Apply $\frac{d}{dx}\ln|u|=\frac{u'}{u}$ with $u=x^{2}-4$ and $u'=2x$:

$y'=\frac{2x}{x^{2}-4}$

Valid wherever $u\ne0$, so on $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$. The bars are what buy you the middle interval, where $x^{2}-4<0$ and $\ln(x^{2}-4)$ would not exist.

$y'(1)=\dfrac{2}{1-4}=-\dfrac23$ and $y'(3)=\dfrac{6}{9-4}=\dfrac65$.

$\ln|y|=5\ln|2x+1|+4\ln|x-3|-3\ln\left(x^{2}+4\right)$; the last factor needs no bars since $x^{2}+4\ge4$.

Differentiate, using $\frac{u'}{u}$ on every term:

$\frac{y'}{y}=\frac{10}{2x+1}+\frac{4}{x-3}-\frac{6x}{x^{2}+4}$

Multiply by $y$ and substitute the original expression back.

At $x=0$: $y(0)=\frac{(1)(81)}{64}=\frac{81}{64}$ and the bracket is $10-\frac43=\frac{26}{3}$, so $y'(0)=\frac{81}{64}\cdot\frac{26}{3}=\frac{351}{32}$.

(a) The power rule is proved for a constant exponent and the $a^{x}$ rule for a constant base. In $x^{x}$ both move, so neither theorem applies; the correct answer is in fact their sum.

(b) $\ln y=x\ln x$, so $\frac{y'}{y}=\ln x+1$ and

$y'=x^{x}(\ln x+1)$

(c) $x^{x}>0$, so $y'=0$ only when $\ln x=-1$, i.e. $x=\frac1e$. Since $y'<0$ on $\left(0,\frac1e\right)$ and $y'>0$ after it, that critical point is the absolute minimum, with value $\left(\frac1e\right)^{1/e}=e^{-1/e}\approx0.6922$.