Special Functions

01Sine and cosine

The two special limits from Module 02 — $\lim_{h\to 0}\frac{\sin h}{h} = 1$ and $\lim_{h\to 0}\frac{1-\cos h}{h} = 0$ — plug into the limit definition to give the cleanest derivative pair in mathematics:

$$\frac{d}{dx}\sin x = \cos x \qquad\qquad \frac{d}{dx}\cos x = -\sin x$$

The hero graph shows why this is believable: wherever $\sin$ peaks or bottoms out (slope zero), $\cos$ crosses zero; wherever $\sin$ climbs steepest, $\cos$ is at its maximum. Keep the sign straight: cosine's derivative carries the minus. These formulas are the reason radians are mandatory — in degrees, an ugly factor of $\pi/180$ contaminates everything.

02The other four trig functions

The remaining four follow from the quotient rule applied to $\sin$ and $\cos$ — derive them once, then memorize:

• $\dfrac{d}{dx}\tan x = \sec^2 x$
• $\dfrac{d}{dx}\sec x = \sec x \tan x$
• $\dfrac{d}{dx}\cot x = -\csc^2 x$
• $\dfrac{d}{dx}\csc x = -\csc x \cot x$

Memory pattern: every "co-" function ($\cos$, $\cot$, $\csc$) gets a negative sign, and each co-derivative is the mirror of its partner with every function swapped for its co-version.

03The exponential function

The number $e \approx 2.71828$ is defined so that this works:

$$\frac{d}{dx}\,e^x = e^x$$

The exponential function is its own derivative — its height is its slope at every point. That self-similarity is why $e^x$ governs everything that grows in proportion to its size: populations, compound interest, radioactive decay. For other bases, a correction factor appears:

$$\frac{d}{dx}\,a^x = a^x \ln a$$

so $\frac{d}{dx}2^x = 2^x\ln 2$. (Check the consistency: when $a = e$, $\ln e = 1$ and the factor vanishes.)

04The natural logarithm

$$\frac{d}{dx}\,\ln x = \frac{1}{x} \qquad (x > 0) \qquad\qquad \frac{d}{dx}\,\log_a x = \frac{1}{x \ln a}$$

A power function's derivative ($x^{-1}$) coming from a non-power function is one of calculus's happiest surprises — it fills the single gap the reverse power rule will have in Module 10. Practical tip: use log laws to expand before differentiating. For $y = \ln\dfrac{x^2\sqrt{x+1}}{(2x-3)^4}$, expanding to $2\ln x + \tfrac{1}{2}\ln(x+1) - 4\ln(2x-3)$ turns a nightmare into three easy chain-rule terms.

05Everything meets the chain rule

Each formula above has a chain-rule version with $u = g(x)$ inside; these generalized forms are what you actually use:

$$\frac{d}{dx}\sin u = \cos u \cdot u' \qquad \frac{d}{dx}\,e^u = e^u \cdot u' \qquad \frac{d}{dx}\ln u = \frac{u'}{u}$$

Examples: $\frac{d}{dx}\sin(5x) = 5\cos(5x)$; $\frac{d}{dx}e^{3x^2} = 6x\,e^{3x^2}$; $\frac{d}{dx}\ln(x^2+1) = \frac{2x}{x^2+1}$. The $\frac{u'}{u}$ pattern for logs is so common it deserves its own neuron. Watch the notation trap: $\sin^2 x$ means $(\sin x)^2$ — a power of sine, differentiated with the chain rule as $2\sin x\cos x$ — while $\sin(x^2)$ has the square inside.

06Inverse trig functions

The two that matter most in Calc 1:

$$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}} \qquad\qquad \frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$$

(Also $\arccos x$, whose derivative is just the negative of arcsine's.) Strange but true: the derivatives of inverse trig functions are purely algebraic. These arrive via implicit differentiation — a preview of Module 07 — and their real importance blooms in integration, where $\frac{1}{1+x^2}$ would otherwise be impossible to antidifferentiate.

07Mixed practice: choosing the pattern

Fluency means recognizing which formula plus which structural rule applies. $y = x^2 e^x$ is a product: $y' = 2xe^x + x^2e^x = xe^x(2 + x)$. $y = e^{\sin x}$ is a composition: $y' = e^{\sin x}\cos x$. $y = \dfrac{\ln x}{x}$ is a quotient: $y' = \dfrac{\frac{1}{x}\cdot x - \ln x}{x^2} = \dfrac{1 - \ln x}{x^2}$. Same six memorized formulas, different scaffolding. When stuck, name the outermost structure aloud — product, quotient, or composition — before writing anything.

08Logarithmic differentiation

For functions with a variable base and variable exponent, like $y = x^x$, neither the power rule (needs constant exponent) nor the exponential rule (needs constant base) applies. The trick: take $\ln$ of both sides, use log laws, then differentiate implicitly. From $\ln y = x\ln x$, differentiating gives $\frac{y'}{y} = \ln x + 1$, so $y' = x^x(\ln x + 1)$. The same technique tames big products and quotients: logs convert multiplication into addition before you differentiate.