Derivatives of Advanced Functions
Trigonometric Functions
In AHL 5.15, students expand their knowledge of derivatives to include more complex trigonometric functions:
- $\frac{d}{dx}(\tan x) = \sec^2 x$
- $\frac{d}{dx}(\sec x) = \sec x \tan x$
- $\frac{d}{dx}(\csc x) = -\csc x \cot x$
- $\frac{d}{dx}(\cot x) = -\csc^2 x$
To find the derivative of $y = \tan(3x)$, we use the chain rule: $\frac{dy}{dx} = \sec^2(3x) \cdot 3 = 3\sec^2(3x)$
Exponential and Logarithmic Functions
For exponential and logarithmic functions with any base $a$:
- $\frac{d}{dx}(a^x) = a^x \ln a$
- $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$
When $a = e$ (Euler's number), these simplify to the familiar forms: $\frac{d}{dx}(e^x) = e^x$ and $\frac{d}{dx}(\ln x) = \frac{1}{x}$
Inverse Trigonometric Functions
The derivatives of inverse trigonometric functions are also covered:
- $\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}$
- $\frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1-x^2}}$
- $\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}$
Do not forget to use the chain rule on these standard derivatives!
- For example
$$ \frac{d}{dx}(a^{e^x})=a^{e^x}\ln(a)\times e^x $$ - Another Example
$$ \frac{d}{dx}(\arctan(x^2))=\frac{1}{1+x^4}(2x) $$
Indefinite Integration
Basic Indefinite Integrals
The indefinite integrals of the functions above are essentially the reverse of their derivatives:
- $\int \sec^2 x , dx = \tan x + C$
- $\int \sec x \tan x , dx = \sec x + C$
- $\int \csc x \cot x , dx = -\csc x + C$
- $\int \csc^2 x , dx = -\cot x + C$
- $\int a^x , dx = \frac{a^x}{\ln a} + C$
- $\int \frac{1}{x \ln a} , dx = \log_a |x| + C$
- $\int \frac{1}{\sqrt{1-x^2}} , dx = \arcsin x + C$
- $\int -\frac{1}{\sqrt{1-x^2}} , dx = \arccos x + C$
- $\int \frac{1}{1+x^2} , dx = \arctan x + C$
Sometimes these functions are incorporated with other linear functions. This often involves a u-substitution.
Example 1:
To integrate $\int \sec^2(2x+5) \, dx$:
Let $u = 2x+5$, then $du = 2dx$ or $dx = \frac{1}{2}du$ therefore
we have $\int \sec^2(2x+5) \, dx = \frac{1}{2}\int \sec^2 u \, du = \frac{1}{2}\tan u + C = \frac{1}{2}\tan(2x+5) + C$
Example 2:
To integrate $\int \frac{1}{x^2+2x+5} \, dx$:
We can write $x^2+2x+5 = x^2+2x+1+4 = 4+(x+1)^2 = 4(1+\frac{(x+1)^2}{4})$ so the integral becomes
$$\int \frac{1}{x^2+2x+5} \, dx =\int \frac{1}{4(1+\frac{(x+1)^2}{4})} \, dx = \frac{1}{4}\int \frac{1}{1+\frac{(x+1)^2}{4}} \, dx$$
Notice that $(\frac{x+1}{2})^2 = \frac{(x+1)^2}{4}$ hence we can let $u = \frac{x+1}{2}$ such that we have $dx = 2\,du$ and the integral becomes
$$\frac{1}{4}\int \frac{1}{1+u^2}2\,du = \frac{1}{2}\int \frac{1}{1+u^2}\,du$$
We can use the standard integral above to obtain
$$\frac{1}{2}(\arctan(u) + C) = \frac{1}{2}\arctan(\frac{x+1}{2}) + C_2$$
Where $\frac{1}{2}C$ is just another constant so we write $C_2$ for simplicity
The example 2 above goes through a complicated $u$-substitution for the integral of $\arctan$ however in the formula booklet, you are given 2 more standard integrals which can make the process easier
- $\int \frac{1}{\sqrt{a^2-x^2}}\,dx=\arcsin(\frac{x}{a}) + C$ where $x^2 < a^2$
- $\int \frac{1}{a^2+x^2}\,dx=\frac{1}{a}\arctan(\frac{x}{a}) + C$
Where (2) can be used in the example above instead of factoring out the number 4.
Assume that you only know $\int \frac{1}{1+x^2}\,dx = \arctan(x)+C$
Try to prove that
$$\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan(\frac{x}{a})+C$$
Indefinite Integral as a Family of Curves
An indefinite integral includes a constant C which we do not know without knowing some information about the actual curve, hence it's indefinite.
This constant shifts the curve vertically without changing its shape depending what the value of the constant is.
Hence the indefinite integral can be thought of as a collection, or family of curves with the same shape but different positions on a vertical axis!
The integral $\int 1\,dx = x + C$ can be many curves/lines such as $x+1$ or $x+1405459$.
Ultimately the derivative of both is the same!
Derivatives are like a summary while the actual function is like a story.
You don't know everything about the actual story if you only read the summary, but you can easily make a summary if you read the actual story!
Partial Fractions in Integration
Partial fractions are a powerful technique for integrating rational functions. This links back to the algebraic methods learned in AHL 1.11.
Sometimes functions are super complicated to integrate, but decomposing them into partial fractions makes integrating easier!
Steps for Partial Fraction Decomposition:
- Ensure the numerator's degree is less than the denominator's. If not, perform long division first.
- Factor the denominator completely.
- Write out the partial fraction decomposition based on the factors.
- Solve for the unknown coefficients.
To integrate $\int \frac{1}{x^2 + 3x + 2} \, dx$:
- Factor the denominator: $x^2 + 3x + 2 = (x+1)(x+2)$
- Set up partial fractions: $\frac{1}{x^2 + 3x + 2} = \frac{A}{x+1} + \frac{B}{x+2}$
- Solve for A and B (multiply both sides by $(x+1)(x+2)$ AHL 1.11)
- We get: $\frac{1}{x^2 + 3x + 2} = \frac{1}{x+1} - \frac{1}{x+2}$
- Now integrate: $\int \frac{1}{x^2 + 3x + 2} \, dx $ $$ \begin{align*} &= \int (\frac{1}{x+1} - \frac{1}{x+2}) \, dx \\ & = \ln|x+1| - \ln|x+2| + C \\ &= \ln|\frac{x+1}{x+2}| + C \end{align*}$$
Hence using partial fractions can be really useful for integration!
When integrating rational functions, always check if partial fraction decomposition is necessary. It can turn a difficult integral into a sum of simpler integrals.
Complex Examples
- $\int e^{3x} \cos(2x) , dx$ This requires integration by parts, possibly twice.
- $\int \frac{x}{x^2 - 4} , dx$ This needs partial fractions followed by basic integration.
