###### Motivating Questions

What other things do we need to know for the exam that were not in our book?

What other things do we need to know for the exam that were not in our book?

- Find the equation of the tangent line to \(f(x)=x^2-3\) at \(x=2\text{.}\) Find the \(x\)-intercept of this tangent line. Sketch \(f(x)\) and the tangent line and mark the \(x\)-intercept of the tangent line.
- Find the equation of the tangent line to \(f(x)=x^2-3\) at \(x=a\text{.}\) Find the \(x\)-intercept of this tangent line.
- Write down the general formula for the tangent line of a function \(f(x)\) at \(x=a\text{.}\) Find the general formula for the \(x\)-intercept of this tangent line.

Solving an equation for \(x\) can be rewritten as solving \(f(x)=0\text{,}\) which means finding the root of \(f(x)\text{.}\) Sometimes we can use algebra to solve for \(x\text{.}\) Other times we can use theory like the Intermediate Value Theorem 1.2.1 to show there is a solution, even though we cannot solve for it. In such situations we can apply a numerical method to find an approximation for the solution.

Newton's Method (also called the Newton-Raphson method) tries to find such an approximate solution by following tangent lines to their \(x\)-intercepts. The idea is:

- Have an approximate solution \(x_i\text{.}\)
- Find the tangent line to \(f(x)\) at \(x=x_i\text{.}\)
- Find the \(x\)-intercept of this tangent line.
- Set \(x_{i+1}\) to be this \(x\)-intercept.

Generally \(x_{i+1}\) will be a better approximate solution than \(x_i\) and repeating this process gives a very good approximation very quickly. Try it in the following computational cell:

In Preview Activity 1.5.1 you figured out the general formula for the \(x\)-intercept of the tangent line. Rather than work out the \(x\)-intercept each time, we can write Newton's method as an iteration using the formula below.

\begin{equation*}
x_{i+1} = x_i -\frac{f(x_i)}{f'(x_i)}
\end{equation*}

- Letting \(f(x)=x^2-3\) and \(x_1=2\text{,}\) use Newton's method to compute \(x_2\text{.}\) Then use Newton's method again to compute \(x_3\text{.}\) If you kept going, what would \(x_i\) converge to?
- What happens if you apply Newton's method to \(f(x)=x^2-3\) using \(x_1=0\text{?}\)
- The function \(f(x)=x^{1/3}\) has \(f'(x)=(1/3)x^{-2/3}\text{,}\) so the Newton's method update forumla gives\begin{equation*} x_{i+1} = x_i -\frac{x_i^{1/3}}{(1/3)x_i^{-2/3}} = x_i - 3 x_i = -2x_i\,. \end{equation*}If you start with \(x_1=1\) and apply Newton's the update repeatedly, what does \(x_i\) converge to?

Sometimes Newton's method fails. If the tangent line is horizontal then it has no \(x\)-intercept and the process breaks. If \(f\) is not differentiable at the root, then the approximations may get worse and worse.

We know

\begin{align*}
\frac{d}{dx} a^x \amp = a^x \ln(a)\quad\text{and}\\
\frac{d}{dx} x^n \amp = nx^{n-1}\,,
\end{align*}

but that does not help help us differentiate \(x^x\text{.}\) To figure out its derivative, we will use a process called logarithmic differentiation.

Set \(y=x^x\text{.}\) Taking the natural logarithm of both sides yields

\begin{equation*}
\ln(y)=\ln\left(x^x\right)\,.
\end{equation*}

Using properties of logarithms we can simplify to

\begin{equation*}
\ln(y)=x \ln(x)\,.
\end{equation*}

This we can differentiate implicitly, using the chain and product rules, to get

\begin{equation*}
\frac{1}{y}\frac{dy}{dx} = 1\ln(x) + x \frac{1}{x}\,.
\end{equation*}

Simplifying and solving for \(\frac{dy}{dx}\) yields

\begin{equation*}
\frac{dy}{dx} = \left(\ln(x)+1\right) y \,.
\end{equation*}

Substituting \(y=x^x\) gives the final answer

\begin{equation*}
\frac{d}{dx} x^x = \left(\ln(x)+1\right)x^x \,.
\end{equation*}

Logarithmic differentiation is required to take the derivative of any function of the form \(f(x)^{g(x)}\) where neither of \(f\) or \(g\) are just constants. It works because it converts \(f(x)^{g(x)}\text{,}\) which we cannot differentiate, into \(g(x) \ln(f(x))\text{,}\) which we can differentiate.

Logarithmic differentiation sometimes gives a quicker way to differentiate functions that we could do by another method. Suppose we want to differentiate

\begin{equation*}
y =
\frac{\left(f(x)\right)^3\left(g(x)\right)^5}{\left(p(x)\right)^7}\,.
\end{equation*}

We could do this using a quotient rule, a product rule, and three chain rules. Using logarithmic differentiation, we compute

\begin{align*}
\ln(y) \amp = \ln\left(\frac{\left(f(x)\right)^3\left(g(x)\right)^5}{\left(p(x)\right)^7}\right)\\
\amp = 3\ln\left(f(x)\right) + 5 \ln\left(g(x)\right) - 7 \ln\left(p(x)\right) \\
\Rightarrow \frac{1}{y}\frac{dy}{dx}\amp = 3 \frac{f'(x)}{f(x)}+5 \frac{g'(x)}{g(x)} -7 \frac{p'(x)}{p(x)} \\
\Rightarrow \frac{dy}{dx}\amp = \left(3 \frac{f'(x)}{f(x)}+5 \frac{g'(x)}{g(x)} -7 \frac{p'(x)}{p(x)}\right) \frac{\left(f(x)\right)^3\left(g(x)\right)^5}{\left(p(x)\right)^7}\,.
\end{align*}

Use logarithmic differentiation to find the derivatives of each of the following.

- \(\displaystyle y = (1+x^2)^{\sin(x)}\)
- \(\displaystyle y = \frac{(1+x^2)^5}{(\sin(x))^4}\)

We learned the derivative of the natural logarithm \(\ln(x)\) in Active Calculus 2.6. Below is the general formula for the derivative of a logarithm.

\begin{equation*}
\frac{d}{dx} \log_a(x) = \frac{1}{x \ln(a)}
\end{equation*}

We learned the derivatives of \(\arcsin(x)\) and \(\arctan(x)\) in Active Calculus 2.6. Below are the rest of the inverse trigonometric functions.

\begin{align*}
\frac{d}{dx} \arcsin(x) \amp =
\frac{d}{dx} \sin^{-1}(x) =
\frac{1}{\sqrt{1-x^2}}\\
\frac{d}{dx} \arctan(x) \amp =
\frac{d}{dx} \tan^{-1}(x) =
\frac{1}{1+x^2}\\
\frac{d}{dx} \mathrm{arcsec}(x) \amp =
\frac{d}{dx} \sec^{-1}(x) =
\frac{1}{x\sqrt{x^2-1}}\\
\frac{d}{dx} \arccos(x) \amp =
\frac{d}{dx} \cos^{-1}(x) = -
\frac{1}{\sqrt{1-x^2}} = - \frac{d}{dx}
\arcsin(x)\\
\frac{d}{dx} \mathrm{arccot}(x) \amp =
\frac{d}{dx} \cot^{-1}(x) = - \frac{1}{1+x^2} =
- \frac{d}{dx} \arctan(x)\\
\frac{d}{dx} \mathrm{arccsc}(x) \amp =
\frac{d}{dx} \csc^{-1}(x) = -
\frac{1}{x\sqrt{x^2-1}} = - \frac{d}{dx}
\mathrm{arcsec}(x)
\end{align*}

In some sources you will find the alternative formula \(\frac{d}{dx} \mathrm{arcsec}(x) = \frac{1}{|x|\sqrt{x^2-1}}\text{.}\) Those sources chose a different domain restriction on \(\sec\) and so a different version of \(\mathrm{arcsec}\text{.}\)

The hyperbolic sine and hyperbolic cosine functions are defined as

\begin{align*}
\sinh(x) \amp = \frac{e^x-e^{-x}}{2}\quad\text{and}\\
\cosh(x) \amp = \frac{e^x+e^{-x}}{2}\,,
\end{align*}

and have the derivatives below.

\begin{align*}
\frac{d}{dx} \sinh(x) \amp = \cosh(x)\\
\frac{d}{dx} \cosh(x) \amp = \sinh(x)
\end{align*}

Note the similarities with \(\sin(x)\) and \(\cos(x)\text{,}\) but there is no negative sign on the derivative of \(\cosh(x)\text{.}\)

Find the derivative of each of the following functions.

- \(\displaystyle f(x) = \log_4(\cosh(3x))\)
- \(\displaystyle g(x) = \mathrm{arccsc}(x)\sinh(x^4)\)
- \(\displaystyle p(x) = \arcsin(x) + \mathrm{arccsc}(x)\)

In Active Calculus 1.8 we learned how to take the formula for the tangent line at \(a\)

\begin{equation*}
y-f(a) =
f'(a)(x-a)
\end{equation*}

and re-interpret it as a local linearization

\begin{equation*}
L(x) = y = f(a) + f'(a)(x-a)\,,
\end{equation*}

which can then be used as a linear approximation \(f(x) \approx L(x)\) when \(x\) is near \(a\text{.}\)

Differentials give a different way to think about the tangent line, local linearization, and linear approximation. Define the differentials

\begin{align*}
dx \amp = x-a \quad\text{and}\\
dy \amp = y-f(a)\,.
\end{align*}

We can then write the tangent line as \(dy = f'(a) dx\text{,}\) which is suggestive of the notation \(\frac{dy}{dx}=f'(x)\) for the derivative. We can rewrite the linear approximation as

\begin{equation*}
f(x)-f(a)\approx L(x)-f(a)=y-f(a)=dy= f'(a)dx\,.
\end{equation*}

Use differentials to estimate the amount of paint (in \(\mathrm{m}^3\)) needed to apply a coat of paint \(0.07\,\mathrm{cm}\) thick to a hemispherical dome with diameter \(60\,\mathrm{m}\text{.}\)

Solution

The volume of a hemisphere is half that of a sphere, and so is \(V=\frac{1}{2}\frac{4}{3}\pi r^3 =\frac{2}{3}\pi r^3\text{.}\) Differentiating with respect to \(r\) gives \(\frac{dV}{dr}=\frac{2}{3}\pi 3 r^2 \) so \(dV=2\pi r^2dr\text{.}\) Inserting the radius of the hemisphere \(r=30\,\mathrm{m}\) and the thickness of the paint \(dr=0.07\,\mathrm{cm}\) yields the paint volume

\begin{equation*}
dV=2\pi
(30\,\mathrm{m})^2(0.07\,\mathrm{cm})=2\pi \cdot
30^2\cdot 0.07\cdot\mathrm{m}^2\cdot\mathrm{cm} =
\frac{2\pi \cdot 30^2\cdot
0.07}{100}\mathrm{m}^3\,.
\end{equation*}

A disc was measured to have diameter \(3\,\mathrm{m}\) with measurement error \(\pm 0.1 \,\mathrm{cm}\text{.}\)

- Use differentials to estimate the error in the calculated area of the disc.
- Use differentials to estimate the relative error in the calculated area of the disc.

In Active Calculus 1.7 we learned that a function being continuous at \(x=a\) means \(\displaystyle \lim_{x\rightarrow a} f(x) = f(a)\text{.}\) Two ways in which a function can fail to be continous are common enough to have their own names:

Graphically a removeable discontinuity is a hole. The name comes from the idea that by redefining \(f(a)\) we could remove the discontinuity and make \(f\) continuous.

Graphically a jump discontinuity is a place where the graph jumps from one value to another.

We learned the Fundamental Theorem of Calculus in Active Calculus 4.4 and the Second Fundamental Theorem of Calculus in Active Calculus 5.2. Other books refer to our Second Fundamental Theorem of Calculus as "The Fundamental Theorem of Calculus Part I" and our Fundamental Theorem of Calculus as "The Fundamental Theorem of Calculus Part II". It is best to ignore the labels and just remember that one allows you to compute a definite integral using antiderivatives and the other allows you to differentiate a function defined as an integral up to \(x\text{.}\)

Use Newton’s method with the specified initial approximation \(x_1\) to find \(x_2\text{,}\) the second approximation to the root of the given equation. Leave the answer as a fraction.

\begin{equation*}
x^7+4=0\quad\text{with}\quad x_1=-1\,.
\end{equation*}

Use logarithmic differentiation to find the derivative of

\begin{equation*}
y = \frac{(1+x^2)^{\sin(x)}(1+x^2)^5}{(\sin(x))^4}\,.
\end{equation*}

Evaluate the definite integral

\begin{equation*}
\int_0^1 \left(3+\cosh(x)\right)dx\,.
\end{equation*}

Find the derivative of

\begin{equation*}
h(x) = \int_0^{\mathrm{arccsc}(x)}\sinh(\log_7(t))\,dt\,.
\end{equation*}

The speed \(v\) of blood flowing along the central axis of an artery of radius \(R\) is given by Poiseulle's law

\begin{equation*}
v(R)=cR^2
\end{equation*}

where \(c\) is a constant. If you can determine the radius of the artery within an accuracy of \(5\%\text{,}\) how accurate is your calculation of the speed?