]> IB HL2 - Chapter 05

Logarithmic, Exponential and Other Transcendental Functions

5.1: The Natural Logarithmic Function: Differentiation

The Definition

First, you must know the real definition of the natural logarithm: $ln x = \int_{1}^{x} \frac{1}{t} d t$ (where $x > 0$ ).

Reminders

${log}_{n} (a \cdot b) = {log}_{n} a + {log}_{n} b$

${log}_{n} (\frac{a}{b}) = {log}_{n} a - {log}_{n} b$

${log}_{n} (a^{b}) = b \cdot {log}_{n} a$

${log}_{n} x = \frac{{log}_{b} x}{{log}_{b} n}$ for any valid base b.

The Number e

Euler's Number; The Natural Number: $e = lim_{x \to \infty} {(1 + \frac{1}{x})}^{x}$ . Also, it is the number so that $\int_{1}^{e} \frac{1}{t} d t$ .

The Derivative of the Natural Logarithmic Function

By the Second Fundamental Theorem of Calculus: $\frac{d}{d x} [ln x] = \frac{1}{x}$ . The chain rule version: $\frac{d}{d x} [ln (f (x))] = \frac{1}{f (x)} f^{'} (x)$ .

Since the domain of the natural log is positive reals, we can sometimes run into trouble…thus, there is another form: $\frac{d}{d x} [ln | x |] = \frac{1}{x}$ . This can help!

Logarithmic Differentiation

Sometimes, the use of logarithms can help you to take derivatives that are otherwise mighty ugly (by taking the log of both sides of the equation). This is because of the fantastic property of logarithms that allows you to bring exponents down as coefficients. Also, the product/quotient properties come in handy.

The only catch is that in using technique, you'll technically need to use implicit differentiation, which means that you might have some y terms that need to be expanded with the original definition…

Examples:

[1.] Find $g^{'} (x)$ if $g (x) = {log}_{10} x$ .

$g (x) = {log}_{10} x = \frac{ln x}{ln 10}$ ⇒ $g^{'} (x) = \frac{1}{ln 10} \cdot \frac{1}{x} = \frac{1}{x \cdot ln 10}$ .

[2.] Find $f^{'} (x)$ if $f (x) = \frac{x}{\sqrt{x^{2} + 1}}$ .

$y = \frac{x}{\sqrt{x^{2} + 1}}$ ⇒ $ln y = ln (\frac{x}{\sqrt{x^{2} + 1}})$ ⇒ $ln y = ln x - ln \sqrt{x^{2} + 1}$ ⇒ $ln y = ln x - \frac{1}{2} ln (x^{2} + 1)$ . Now take the derivative! $\frac{1}{y} y^{'} = \frac{1}{x} - \frac{1}{2} (\frac{1}{x^{2} + 1}) (2 x)$ . Solve and substitute: $y^{'} = y (\frac{1}{x} - \frac{x}{x^{2} + 1}) = \frac{x}{\sqrt{x^{2} + 1}} (\frac{1}{x} - \frac{x}{x^{2} + 1}) = \frac{1}{\sqrt{x^{2} + 1}} - \frac{x^{2}}{{(x^{2} + 1)}^{\frac{3}{2}}}$ .

5.2: The Natural Logarithmic Function: Integration

The Integrals

Just work backwards! $\int \frac{1}{x} d x = ln | x | + C$ . Naturally, life is not always this simple; perhaps a version ready-made for substitution is on order: $\int \frac{u^{'}}{u} d u = ln | u | + C$ .

There is not, however, an easy solution to $\int ln x d x$ . Perhaps we'll chat about that later…

Integrals of Trigonometric Functions

The integrals of sine and cosine are easy; what about tangent?

$\int tan x d x = \int \frac{sin x}{cos x} d x = - \int \frac{- sin x}{cos x} d x = - \int \frac{u^{'}}{u} d u$ , if you use $u = cos x$ . Thus, $\int tan x d x = - ln | cos x | + C$ . You can use a similar method to find…

Examples

[3.] $\int \frac{1}{3 x + 2} d x = ?$

I notice that this is "one over something," and (naturally) think of logarithms. If $u = 3 x + 2$ , then $d u = 3 d x$ . Thus, $\int \frac{1}{3 x + 2} d x = \int \frac{1}{u} \frac{d u}{3} = \frac{1}{3} \int \frac{d u}{u}$ . Aha! $\frac{1}{3} \int \frac{d u}{u} = \frac{1}{3} ln | u | + C = \frac{1}{3} ln | 3 x + 2 | + C$ .

[4.] $\int \frac{1}{x \cdot ln x^{3}} d x = ?$

Sneaky…at first glance, there are several things that I could use for u…however, leaving the log in the denominator isn't an option; that must be part of my u. $u = ln x^{3} ⇒ d u = \frac{1}{x^{3}} \cdot 3 x^{2} d x ⇒ d u = \frac{3}{x} d x$ . So $\int \frac{1}{x \cdot ln x^{3}} d x = \int \frac{1}{x} \cdot \frac{1}{u} \cdot d x = \int \frac{1}{u} \cdot \frac{d x}{x} = \int \frac{1}{u} . \frac{d u}{3} = \frac{1}{3} \int \frac{d u}{u}$ . Finally, $\frac{1}{3} \int \frac{d u}{u} = \frac{1}{3} ln | u | + C = \frac{1}{3} ln | ln x^{3} | + C$ .

[5.] $\int sec (\frac{x}{2}) d x = ?$

Easy! $\int sec (\frac{x}{2}) d x = 2 \int sec u d u = 2 ln | sec u + tan u | + C = 2 ln | sec (\frac{x}{2}) + tan (\frac{x}{2}) | + C$ .

[6.] $\int \frac{{csc}^{2} x}{cot x} d x = ?$

Another tricky one…however, note that the derivative of the denominator is almost the same as the numerator. $u = cot x ⇒ d u = - {csc}^{2} x d x$ . So $\int \frac{{csc}^{2} x}{cot x} d x = \int \frac{- d u}{u} = - ln | u | + C = - ln | cot x | + C$ .

5.4: Exponential Functions: Differentiation and Integration

The Natural Exponential Function

First, the real definition: if $f (x) = ln x$ , then $e^{x} = f^{- 1} (x)$ . In other words, $y = e^{x} ⇔ x = ln y$ . The Natural Exponential function is defined as the inverse function of the natural logarithmic function.

Reminders

$e^{m} e^{n} = e^{m + n}$

$\frac{e^{m}}{e^{n}} = e^{m - n}$

Don't forget that the domain of the natural log is the range of the natural exponential, etc.

Derivatives of Exponential Functions

Since $ln e^{x} = x$ , $\frac{d}{d x} [ln e^{x}] = \frac{d}{d x} [x]$ . This becomes $\frac{1}{e^{x}} \frac{d}{d x} [e^{x}] = 1 ⇒ \frac{d}{d x} [e^{x}] = e^{x}$ !

If you want to take the derivative of an exponential function other than the natural, you'll have to use logarithmic differentiation…more on that in the next section!

Integrals of Exponential Functions

Easy! $\int e^{x} d x = e^{x} + C$ .

Examples

[7.] Find the gradient of the line tangent to $y = e^{- 2 x}$ at the point $(- 1, e^{2})$ .

First, the derivative: $y^{'} = - 2 e^{- 2 x}$ . At the indicated point, the gradient is $- 2 e^{- 2} = - \frac{2}{e^{2}}$ .

[8.] $\int \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x = ?$

Clearly, this requires a substitution…but which piece? I suspect the denominator. $u = e^{x} + e^{- x} ⇒ d u = (e^{x} - e^{- x}) d x$ . So $\int \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x = \int \frac{d u}{u} = ln | u | + C = ln | e^{x} + e^{- x} | + C$ .

5.5: Bases Other than e

Definitions

$a^{x} = e^{x \cdot ln a}$ for any base $a \in (0, 1) \cup (1, \infty)$ .

${log}_{a} x = \frac{ln x}{ln a} = (\frac{1}{ln a}) ln x$ for any valid base a.

Differentiation and Integration

$\frac{d}{d x} [a^{x}] = (ln a) (a^{x})$ , and $\int a^{x} d x = (\frac{1}{ln a}) a^{x} + C$ .

$\frac{d}{d x} [{log}_{a} x] = \frac{1}{(ln a) (x)}$

Examples

[9.] Find y′ if y = 2 ^x .

First, take the log of both sides: $ln y = ln 2^{x}$ . Now pull that exponent down: $ln y = x \cdot ln 2$ . Remember that $ln 2$ is a constant! Take the derivative, remembering that you are really doing implicit differentiation: $\frac{1}{y} y^{'} = ln 2$ . Solve for y′: $y^{'} = y \cdot ln 2$ . Finally, replace the y with its original definition—in this case, y = 2 ^x : $y^{'} = 2^{x} \cdot ln 2$ . Or, use the definition above.

[10.] Find $f^{'} (x)$ if $f (x) = {log}_{10} x^{2}$ .

Change of base, anyone? $f (x) = \frac{ln x^{2}}{ln 10}$ , so $f^{'} (x) = \frac{1}{ln 10} \cdot \frac{1}{x^{2}} \cdot 2 x = \frac{2}{xln 10}$ . Or, use the formula…

[11.] $\int 2^{x} d x = ?$

If $u = 2^{x}$ , then $d u = (ln 2) (2^{x}) d x$ . So $\int 2^{x} d x = \int \frac{d u}{ln 2}$ . Keep going! $\frac{1}{ln 2} \int d u = \frac{1}{ln 2} \cdot u + C = \frac{1}{ln 2} 2^{x} + C$ . Or…you know.

5.7: Differential Equations: Separation of Variables

Definition

A differential equation is an equation that involves derivatives. Sounds simple, eh?

General and Particular Solutions

The solution to a differential equation is a function that, when it and its derivatives are plugged into the differential equation, causes the equation to be true. Sounds complicated, eh?

Working from a derivative to a function will (naturally) involve integration. Since indefinite integration always creates a constant C, we may obtain general solutions (families of solutions that differ only in the value of the constant). As we have seen previously, if some information (initial condition) is given, then we can find a particular solution.

Differential equations are classified according to the highest derivative that appears in the equation—this is called the order of the differential equation.

Separation of Variables

If it is possible to write a differential equation so that every term involving x is on one side of the equation, and every term involving y is on the other, then the differential equation is separable. As a result, the equation can be solving by separating the variables and integrating.

Homogenous Differential Equations

Some non-separable differential equations can be made separable through a change of variables…however, this is more difficult, and not part of the IB syllabus (nor the AP, as best I can tell…), so we won't consider it.

Examples

[12.] Find the general solution to $\frac{d y}{d x} = x^{3} - 4 x$ .

Well—the variables are already separated! Let's just write it in differential form: $d y = (x^{3} - 4 x) d x$ . Now, integrate both sides: $\int d y = \int (x^{3} - 4 x) d x ⇒ y = \frac{1}{4} x^{4} - 2 x^{2} + C$ . Note that you only need one constant—the two that you would normally get can be joined together into one.

[13.] Find the general solution to $x y^{'} = y$ .

First, let's change notation, and put things in differential form: $x \cdot d y = y \cdot d x$ . Now, separate the variables: $\frac{1}{y} \cdot d y = \frac{1}{x} \cdot d x$ . Integrate: $ln | y | = ln | x | + C$ . Now, solve for y: $| y | = e^{ln | x | + C} = e^{C} e^{ln | x |} = e^{C} | x |$ . Remember that absolute value can be removed with a plus-minus symbol. $y = \pm e^{C} | x |$ .

[14.] Find the particular solution to $2 x \frac{d y}{d x} = ln x^{2}$ if $y (1) = 2$ .

Well, let's find the general solution first: $\frac{d y}{d x} = \frac{ln x^{2}}{2 x} = \frac{2 ln | x |}{2 x} = \frac{ln | x |}{x}$ . Now, differential form: $d y = \frac{ln | x |}{x} d x$ . Integrating the right side requires a substitution: $u = ln | x | ⇒ d u = \frac{1}{x} d x$ . Thus, the differential can be written $d y = u \cdot d u$ , which integrates to $y = \frac{1}{2} u^{2} + C$ . Undo the substitution to get the general solution: $y = \frac{1}{2} {(ln | x |)}^{2} + C$ .

Now, for the particulars: $y (1) = 2$ is a way of saying that the point $(1, 2)$ works; thus, $2 = \frac{1}{2} {(ln | 1 |)}^{2} + C ⇒ 2 = \frac{1}{2} {(0)}^{2} + C ⇒ 2 = C$ . So, the particular solution is $y = \frac{1}{2} {(ln | x |)}^{2} + 2$ .

5.8: Inverse Trigonometric Functions: Differentiation

Definitions

Recall that in general, none of the trigonometric functions have inverses, since none of them are one-to-one.

However, also recall that we (you!) previously restricted the domains of some of these functions in order to create inverse functions.

Inverse Function	Domain	Range
arcsine	[-1, 1]	$[- \frac{π}{2}, \frac{π}{2}]$
arccosine	[-1, 1]	[0, π]
arctangent	$ℝ$	$[- \frac{π}{2}, \frac{π}{2}]$

Our text goes so far as to also create inverse function for the other three trigonometric functions! Of course, there is rarely such a need…especially since they are not covered in the IB curriculum.

Derivatives

You can derive the derivatives of the inverse trigonometric functions through implicit differentiation.

$\frac{d}{d x} [arcsin u] = \frac{u^{'}}{\sqrt{1 - u^{2}}}$ , $\frac{d}{d x} [arccos u] = - \frac{u^{'}}{\sqrt{1 - u^{2}}}$ , and $\frac{d}{d x} [arctan u] = \frac{u^{'}}{1 + u^{2}}$ .

Examples

[15.] Find $\frac{d}{d x} [arctan \sqrt{x}]$ .

Basic rule and chain rule…or just follow the formula above. $\frac{d}{d x} [arctan \sqrt{x}] = \frac{\frac{1}{2 \sqrt{x}}}{1 + | x |} = \frac{1}{(2 \sqrt{x}) (1 + | x |)}$ .

[16.] Find $f^{'} (x)$ if $f (x) = x \cdot arcsin x$ .

Product rule? $f^{'} (x) = x \cdot \frac{1}{\sqrt{1 - x^{2}}} + arcsin x$ .

5.9: Inverse Trigonometric Functions: Integration

Of course, some functions, when integrated, give the inverse trigonometric functions!

$\int \frac{1}{\sqrt{a^{2} - u^{2}}} d u = arcsin \frac{u}{a} + C$ , $\int - \frac{1}{\sqrt{a^{2} - u^{2}}} d u = arccos \frac{u}{a} + C$ , and $\int \frac{1}{a^{2} + u^{2}} d u = \frac{1}{a} arctan \frac{u}{a} + C$ .

The difference in the integrals of arcsine and arccosine are such that arcsine is almost always used.

As was the case with natural logarithms, finding $\int arcsin x d x$ is much harder, and will come later…

Examples

[17.] Find $\int \frac{3}{\sqrt{1 - 4 x^{2}}} d x$ .

This looks a lot like the pattern for arcsine…but with a substitution. $u = 2 x ⇒ d u = 2 d x$ . Thus, $\int \frac{3}{\sqrt{1 - 4 x^{2}}} d x = \int \frac{3}{\sqrt{1 - u^{2}}} \frac{d u}{2} = \frac{3}{2} \int \frac{1}{\sqrt{1 - u^{2}}} d u$ . Continue: $\frac{3}{2} \int \frac{1}{\sqrt{1 - u^{2}}} d u = \frac{3}{2} arcsin u + C = \frac{3}{2} arcsin (2 x) + C$ .

[18.] Find $\int \frac{cos x}{1 + {sin}^{2} x} d x$ .

This looks like it might be an arctangent problem… $u = sin x ⇒ d u = cos x d x$ . Thus, $\int \frac{cos x}{1 + {sin}^{2} x} d x = \int \frac{1}{1 + u^{2}} d u$ . Aha! $\int \frac{1}{1 + u^{2}} d u = arctan u + C = arctan (sin x) + C$ .

Page last validated 2010-08-15