Both methods yield \(\fe{\fd{f}}{x}=\frac{3}{2\sqrt{3x+1}}\).

First Method:

\begin{align*} \fe{\fd{f}}{x}&=\lim_{a\to x}\frac{\fe{f}{x}-\fe{f}{a}}{x-a}\\ &=\lim_{a\to x}\frac{\sqrt{3x+1}-\sqrt{3a+1}}{x-a}\\ &=\lim_{a\to x}\frac{\sqrt{3x+1}-\sqrt{3a+1}}{x-a}\cdot\frac{\sqrt{3x+1}+\sqrt{3a+1}}{\sqrt{3x+1}+\sqrt{3a+1}}\\ &=\lim_{a\to x}\frac{(3x+1)-(3a+1)}{(x-a)\left(\sqrt{3x+1}+\sqrt{3a+1}\right)}\\ &=\lim_{a\to x}\frac{3(x-a)}{(x-a)\left(\sqrt{3x+1}+\sqrt{3a+1}\right)}\\ &=\lim_{a\to x}\frac{3}{\sqrt{3x+1}+\sqrt{3a+1}}\\ &=\frac{3}{\sqrt{3x+1}+\sqrt{3x+1}}\\ &=\frac{3}{2\sqrt{3x+1}} \end{align*}

Second Method:

\begin{align*} \fe{\fd{f}}{x}&=\lim_{h\to0}\frac{\fe{f}{x+h}-\fe{f}{x}}{h}\\ &=\lim_{h\to0}\frac{\sqrt{3(x+h)+1}-\sqrt{3x+1}}{h}\\ &=\lim_{h\to0}\frac{\sqrt{3(x+h)+1}-\sqrt{3x+1}}{h}\cdot\frac{\sqrt{3(x+h)+1}+\sqrt{3x+1}}{\sqrt{3(x+h)+1}+\sqrt{3x+1}}\\ &=\lim_{h\to0}\frac{\left(3(x+h)+1\right)-(3x+1)}{h\left(\sqrt{3(x+h)+1}+\sqrt{3x+1}\right)}\\ &=\lim_{h\to0}\frac{3h}{h\left(\sqrt{3(x+h)+1}+\sqrt{3x+1}\right)}\\ &=\lim_{h\to0}\frac{3}{\left(\sqrt{3(x+h)+1}+\sqrt{3x+1}\right)}\\ &=\frac{3}{\sqrt{3x+1}+\sqrt{3x+1}}\\ &=\frac{3}{2\sqrt{3x+1}} \end{align*}

\(1+\dfrac{1}{\sqrt{x^3}}\)

\begin{align*} \fe{f}{x}&=\frac{x^{2} - 2 \, \sqrt{x}}{x}\\ \text{Expand this to:}\\ \fe{f}{x}&=x - \frac{2}{\sqrt{x}}\\ \text{Differentiate each part using the power law: } \de {x^r}x = r x^{r-1}\\ \fe{\fd{f}}{x}&=1+\dfrac{1}{\sqrt{x^3}} \end{align*}

\(9.9000\)

\begin{align*} \fe{f}{x}&=\fe{f}{a} + \fe{\fd{f}}{a}(x-a)\\ TBD \end{align*}

\(\frac{dy}{dx} = -\frac{y \cos\left(x y\right)}{x \cos\left(x y\right) - \cos\left(x\right)} - \frac{y \sin\left(x\right)}{x \cos\left(x y\right) - \cos\left(x\right)}\)

\begin{align*} y \cos\left(x\right)&=1 + \sin\left(x y\right)\\ \text{Differentiate through:}\\ y \frac{d\left(\cos\left(x\right)\right)}{dx} + \frac{dy}{dx}\cos\left(x\right)&= \frac{d\left(\sin\left(xy\right)\right)}{dx}\\ - y \sin\left(x\right) + \frac{dy}{dx}\cos\left(x\right)&= \frac{d\left(\sin\left(xy\right)\right)}{dx} \\ \text{Now set } u = xy\\ \frac{d\left(\sin\left(u\right)\right)}{dx}&= \frac{d\left(\sin\left(u\right)\right)}{du}\frac{du}{dx}\\ &= \cos\left(u\right)\frac{du}{dx}\\ &= \cos\left(u\right)\left(x\frac{dy}{dx} + y\frac{dx}{dx}\right)\\ &= x\cos\left(xy\right)\frac{dy}{dx} + y\cos\left(xy\right)\\ \text{Substituting this into our main equation:}\\ - y \sin\left(x\right) + \frac{dy}{dx}\cos\left(x\right)&= x\cos\left(xy\right)\frac{dy}{dx} + y\cos\left(xy\right) \\ \text{Rearranging:}\\ \frac{dy}{dx}&= \frac{y \cos\left(x y\right) + y \sin\left(x\right)}{ \cos\left(x\right) - x \cos\left(x y\right) } \end{align*}

\(\fe{\fd{y}}{x} = \frac{y}{x} - \frac{2}{x} - 6\)

Implicit: \begin{gather*} xy+2x+3x^2 = 4\\ \text{Differentiate each part using the product rule for } xy\\ x\frac{dy}{dx} + y\frac{dx}{dx} + 2 + 6x = 0\\ x\frac{dy}{dx} + y + 2 + 6x = 0\\ \text{Rearranging gives:} \\ \frac{dy}{dx} = -\frac{\left(y+2+6x\right)}{x}\\ \text{Rearranging original equation for } y\\ y = \frac{4}{x} -2 -3x\\ \text{Substituting gives } y\\ \frac{dy}{dx} = -\frac{\left(\frac{4}{x} -2 -3x+2+6x\right)}{x}\\ \frac{dy}{dx} = -\frac{4}{x^2}-3 \end{gather*}

Explicit: \begin{gather*} xy+2x+3x^2 = 4\\ \text{Solve for } y\\ y = \frac{4}{x} -2 -3x\\ \text{And differentiate } y\\ \frac{dy}{dx} = -\frac{4}{x^2} -3 \end{gather*}

As you can see the results are the same for each method.

The problem that implicit differentiation solves is that it is often difficult or impossible to rearrange to have \(y\) on the left by its own.

After a long period of time everyone knows the rumour!

\begin{align*} lim_{t \to \infty}\fe{p}{t}&=\frac{1}{1+ a e^{-k\infty}}\\ \text{Since }e^{-k\infty} = 0, \{ k \in \mathbb{R} | k > 0 \}\\ \therefore lim_{t \to \infty} p(t)&= \frac{1}{1+0}\\ &=1 \end{align*}

\begin{align*} u&= 1 + a e^{-kt}\\ \frac{dp}{dt}&=\frac{dp}{du}\frac{du}{dt}&\text{(Chain Rule)}\\ &=\frac{-1}{u^2}\left(-ak e^{-kt}\right)\\ &=\frac{ak e^{-kt}}{(1+a e^{-kt})^2} \end{align*}

\begin{align*} p&= \frac{1}{1+ a e^{-kt}}\\ \text{Substitute p's for t's, and vice versa:}\\ t&= \frac{1}{1+ a e^{-kp}}\\ \text{Solve for p:}\\ 1 + a e^{-kp}&=\frac{1}{t}\\ a e^{-kp}&=\frac{1}{t} - 1\\ e^{-kp}&=\frac{1}{at} - \frac{1}{a}\\ -kp&=\ln \left(\frac{1}{at} - \frac{1}{a}\right)\\ p&=-\frac{1}{k}\ln \left(\frac{1-t}{at}\right)\\ p&=\frac{1}{k}\ln \left(\frac{at}{1-t}\right)\\ \text{In other words the inverse is:}\\ p^{-1}&=\frac{1}{k}\ln \left(\frac{at}{1-t}\right) \end{align*}

\(7.38\) days.

See Sage plot  of \(p(t)\) below.

Use the inverse function with a value of \(0.8\) \begin{align*} p^{-1}&=\frac{1}{k}\ln \left(\frac{at}{1-t}\right)\\ p^{-1}(0.8)&=7.378 \end{align*}

See Sage plot  of inverse \(p(t)\) below.

The mass (in \(mg\)) after time \(t\) (in days) is: \begin{equation*}M_t=800 e^{-0.1386t}\end{equation*}

The rate of change of mass is proportional to the current mass: \begin{equation*}\frac{dM}{dt} \propto M\end{equation*} Lets call the constant of proportionality \(\lambda\). \begin{equation*}\frac{dM}{dt} = -\lambda M\end{equation*} The negative is because this is a decay.

The solution of this equation is: \begin{equation*}M_t = M_0 e^{-\lambda t}\end{equation*} (where \(M_0\) is the mass at \(t=0\).)

At the half-life \(t=T_{\frac{1}{2}}\) and \(M_t = \frac{M_0}{2}\): \begin{equation*}\frac{M_0}{2} = M_0 \exp\left({-\lambda T_{\frac{1}{2}}}\right)\end{equation*} \begin{equation*}2 = \exp\left({\lambda T_{\frac{1}{2}}}\right)\end{equation*} \begin{equation*}\log{2} = \lambda T_{\frac{1}{2}}\end{equation*} \begin{equation*}\lambda = \frac{log2}{T_{\frac{1}{2}}}\end{equation*}

We are told that the half-life is \(5\) days. \begin{equation*}\lambda \approx 0.1386\end{equation*}

Substituting with the original mass gives us: \begin{equation*}M_t = 800 e^{-0.1386t}\end{equation*} where \(M_t\) is the mass in \(mg\) and \(t\) is the time in days.

\(N(30) = 12.5 mg\)

\begin{equation*}N(30) = 800 e^{-0.1386 \times 30} = 800 e^{-4.1589} = 800 \times 0.015625 = 12.5 \end{equation*}

\(48.2\) days.

\begin{align*} M_t&=M_0 e^{-\lambda t}\\ 1&=800e^{-\lambda t}\\ \frac{1}{800}&=e^{-\lambda t}\\ 800&=e^{\lambda t}\\ log{800}&=\lambda t\\ t&=\frac{log{800}}{\lambda}\\ &=\frac{log{800}}{0.1386}\\ & \approx 48.2 \end{align*}

See Sage plot  of mass function below.

The first derivative is given by: \begin{equation*}\frac{\Delta f(x)}{\Delta x} = \lim_{h \to 0}\frac{f(x+h)-f(x)}{(x+h)-(x)} = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\end{equation*}

The solution is to apply this twice. The trick here is to start with the correct increments (see Math Stack Q&A [3.7.3.1.1]).

\begin{align*} \fe{f''}{x}&=\lim_{h \to 0} \frac{\fe{f'}{x} - \fe{f'}{x-h}}{h}\\ &=\lim_{h \to 0}\frac{\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}}{h}\\ &=\lim_{h \to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2} \end{align*}

First set \(a=x+h\): \begin{equation*}f(x)f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots\end{equation*} TBD

\begin{align*} \text{Apply the mean value theorem}\\ &f'(c)=\frac{f(8) - f(2)}{8-2}\\ \text{Since we know}\\ &3 \leq \fe{f'}{x} \leq 5\\ \text{We get}\\ &3 \leq \frac{f(8) - f(2)}{6} \leq 5\\ &18 \leq f(8) - f(2) \leq 30 \end{align*}

Remind yourself how to obtain the homogeneous (null) solution:

t,a = var('t,a')
y = function('y')(t)
de1 = diff(y,t) == a*y
sol1 = desolve(de1,y, ivar=t)
show(sol1)

\(C e^{\left(a t\right)}\) and is the general solution to \(\frac{dy}{dt} - ay = 0\).

Also have a look at the screenshot above!

Sage Commands:

t, s, a, y0 = var ('t, s, a, y0')
y = function('y')(t)
de = diff(y,t) == a*y + e^(s*t)
sol=desolve(de,y, ivar=t, ics=[0,y0])
show(sol)

This results in:

This is not quite the same format as Strang but it is easy to see that it is equivalent.

As mentioned above this solution can be taken further by using L'Hopital's Rule.4.0.3 We can circumvent the \(0/0\) catastrophe by differentiating the numerator and denominator.

Sage Commands:

numerator = e^(s*t) - e^(a*t)
numerator.diff(s)
denominator= s - a
denominator.diff(s)

This results in \(\frac{f'(x)}{g'(x)}=\frac{t e^{s t}}{1}=t e^{s t}\)

Hence the full solution using L'Hopital now matches Strang's final solution:

Sage Commands:

t, omega, a, y0 = var ('t, omega, a, y0')
y = function('y')(t)
de = diff(y,t) == a*y + cos(omega*t)
sol=desolve(de,y, ivar=t, ics=[0,y0])
show(sol)

This results in: