Class 4
Definition:
\[\ln x = \int_1^x \frac{1}{t}\; dt\]
Domain: \((0,\infty)\)
\(\ln 1 = 0\)
\((\ln x)' = \frac{1}{x}\)
\(\ln\) is continuous, increasing and concave down.
\(\displaystyle \lim_{x\to\infty} \ln x = \infty\)
\(\displaystyle \lim_{x\to 0^+} \ln x = -\infty\)
\(\ln (xy) = \ln x + \ln y\)
If \(r\) is rational, then \(\ln \left(x^r\right) = r\ln x\).
\(\ln\) is continuous and increasing \(\Longrightarrow\) one-to-one.
It has an inverse function.
We will call it the natural exponential function, denoted as \(\exp()\).
Domain of \(\exp\) is the range of \(\ln\): \((-\infty, \infty)\).
Range of \(\exp\) is the domain of \(\ln\): \((0,\infty)\).
\(\exp(\ln x) = x\) for any \(x \in (0,\infty)\).
\(\ln \exp(y) = y\) for ant \(y \in (-\infty, \infty)\).
What is the derivative of \(\exp\)?
Theorem: If a function \(f\) is a one-to-one function with inverse \(f^{-1}\), and \(y\) is a number such that:
then \(f^{-1}\) is differentiable at \(y\) and \[\left(f^{-1}\right)'(y) = \frac{1}{f'\left(f^{-1}\left(y\right)\right)}\]
Let \(f(x) = 2x^3 + 3x^2 + 7x + 4\). Find \(\left(f^{-1}\right)'(-2)\).
Let \(f(x) = x^3 + 3\sin x + 2\cos x\). Find \(\left(f^{-1}\right)'(2)\).
Let \(f(x) = \ln x\). Find \(\left(f^{-1}\right)'(y) = \exp'(y)\).
\[\begin{align} \left(f^{-1}\right)'(y) &= \frac{1}{f'\left(f^{-1}\left(y\right)\right)} \\[.8em] &\class{fragment}{{}= \frac{1}{\ln'\left(\exp\left(y\right)\right)}} \\[.8em] &\class{fragment}{{}= \frac{1}{\frac{1}{\exp(y)}}} \\[.8em] &\class{fragment}{{}= \color{red}{\exp(y)}} \end{align}\]
The exponential function is differentiable everywhere and
\[\exp'(x) = \exp(x)\]
Since \(\exp'(x) > 0\), the natural exponential function is increasing.
Since \(\exp'(x)\) is increasing, the natural exponential function is concave up.
Define \(e = \exp(1)\), so that \(\ln e = 1\).
Suppose \(r\) is a rational number. Then for any \(x > 0\), \(\ln x^r = r\ln x\).
Applying the exponential function on both sides: \(\exp\left(\ln x^r\right) = \exp\left(r\ln x\right)\).
Since \(\exp\) and \(\ln\) are inverse to each other, they cancel: \(\exp\left(\ln x^r\right) = x^r\).
This gives us \(x^r = \exp\left(r \ln x\right)\).
If \(x = e\), then \(\ln x = 1\), which gives us \(\color{red}{e^r = \exp(r)}\).
For a rational number \(r = \frac{p}{q}\), \(\exp(r) = e^r = \sqrt[q]{e^p}\).
If \(x\) is an irrational number, we define \(e^x\) to be \(\exp(x)\).
Then \(\exp(x) = e^x\) for any real number \(x\).
General powers:
Suppose \(x\) and \(y\) are real numbers, and \(x > 0\). Define \(x^y = \exp(y\ln x)\).
Let \(a\) and \(b\) be two real numbers. Let \(c = \exp(a)\exp(b)\).
\[\begin{align} \class{fragment}{\ln c}& \class{fragment}{{}=\ln\left(\exp(a)\exp(b)\right)} \\[.5em] \class{fragment}{\ln c}& \class{fragment}{{}=\ln\left(\exp(a)\right) + \ln\left(\exp(b)\right)} \\[.5em] \class{fragment}{\ln c}& \class{fragment}{{}=a + b} \\[.5em] \class{fragment}{c}& \class{fragment}{{}=\exp(a + b)} \\[.5em] \class{fragment}{\exp(a+b)}& \class{fragment}{{}=\exp(a)\exp(b)} \end{align}\]
Let \(y = \ln x\), or \(x = \exp(y)\).
\(y\) is an increasing function of \(x\), and \(x\) is an increasing function of \(y\).
As \(x \to \infty\), \(y \to \infty\).
Therefore, as \(y \to \infty\), \(x \to \infty\).
As \(x \to 0^+\), \(y \to -\infty\).
Therefore, as \(y \to -\infty\), \(x \to 0^+\).
\(\displaystyle \lim_{y\to \infty} \exp(y) = \infty\)
\(\displaystyle \lim_{y\to -\infty} \exp(y) = 0\)
Domain of \(\exp\) is the range of \(\ln\): \((-\infty, \infty)\).
Range of \(\exp\) is the domain of \(\ln\): \((0,\infty)\).
\(\exp(\ln x) = x\) for any \(x \in (0,\infty)\).
\(\ln \exp(y) = y\) for ant \(y \in (-\infty, \infty)\).
\(\exp'(y) = \exp(y)\)
\(\exp\) is increasing and concave up.
\(\displaystyle \lim_{y\to \infty} \exp(y) = \infty\)
\(\displaystyle \lim_{y\to -\infty} \exp(y) = 0\)
\(\exp(x + y) = \exp(x)\exp(y)\)
\(\exp(-y) = \frac{1}{\exp(y)}\)
\(f(t) = \tan t\), \(f^{-1}(z) = \tan^{-1}(z) = \arctan z = \operatorname{atan} z\).
\(\displaystyle \arctan' z = \frac{1}{\tan' \left(\arctan z\right)}\class{fragment}{{}= \frac{1}{\sec^2\left(\arctan z\right)}}\)
\(\displaystyle\ln' x = \frac{1}{x}\)
\(\exp'(x) = \exp(x)\)
\(\displaystyle\arcsin' y = \frac{1}{\sqrt{1 - y^2}}\)
\(\displaystyle\arccos' x = -\frac{1}{\sqrt{1 - x^2}}\)
\(\displaystyle\arctan' z = \frac{1}{1 + z^2}\)
\(\displaystyle\operatorname{arccot}' w = -\frac{1}{1 + w^2}\)
\(\displaystyle \int \frac{1}{x}\;dx = \begin{cases} \ln(-x) + C_1 & \text{if } x < 0\\ \ln x + C_2 & \text{if } x > 0\end{cases}\)
\(\displaystyle \int \exp(x)\;dx = \exp(x) + C\)
\(\displaystyle \int \frac{1}{\sqrt{1 - y^2}}\;dy = \arcsin y + C_1 = -\arccos y + C_2\)
\(\displaystyle \int \frac{1}{1 + z^2}\;dz = \arctan z + C_1 = -\operatorname{arccot} z + C_2\)