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Calculus BC · Unit 2 · 2.7
AP Calculus · Differentiation: Definition and Fundamental Properties
Derivatives of $\cos x$, $\sin x$, $e^{x}$ and $\ln x$
Four results to quote from memory — and one minus sign that decides most of the marks.
key idea$\sin\to\cos\to-\sin\to-\cos\to\sin;\qquad \left(e^{x}\right)'=e^{x}$
The rule
Four derivatives to know cold. They are not derived on the exam — they are quoted. The only real difficulty is the minus sign on $\cos$ and the domain restriction on $\ln$.
| $f(x)$ | $f'(x)$ | Worth noting |
|---|
| $\sin x$ | $\cos x$ | no sign change |
| $\cos x$ | $-\sin x$ | minus — the one people drop |
| $e^{x}$ | $e^{x}$ | its own derivative — unique among functions |
| $\ln x$ | $\dfrac{1}{x}$ | valid only for $x\gt 0$ |
| $a^{x}$ | $a^{x}\ln a$ | reduces to $e^{x}$ when $a=e$, since $\ln e=1$ |
| $\log_{a}x$ | $\dfrac{1}{x\ln a}$ | reduces to $\tfrac1x$ when $a=e$ |
The last two rows generalise the middle two. If you remember $a^{x}\ln a$, you never need $e^{x}$ separately.
The sine cycle$\sin\to\cos\to-\sin\to-\cos\to$ back to $\sin$ — period 4
Why $e$ matters$e$ is defined as the base where the slope equals the height
Radians onlythese derivatives are false in degrees
Pattern recognition
structural facts
- Differentiating $\sin$ or $\cos$ four times returns the original — use mod 4 for high orders.
- $e^{x}$ is unchanged by differentiation, so $e^{x}$ terms survive every order.
- $\ln x$ turns a transcendental into a rational function — the only one of the four that changes type.
- All four assume the argument is exactly $x$; anything else needs the chain rule.
Worked examples
EX 1
Straight application
$3\sin x-2\cos x$Differentiate $f(x)=3\sin x-2\cos x$.
1
Differentiate each term
$\dfrac{d}{dx}[\sin x]=\cos x$ and $\dfrac{d}{dx}[\cos x]=-\sin x$.
2
Keep the coefficients
$3\cos x$ and $-2\cdot(-\sin x)=+2\sin x$.
3
Assemble
$f'(x)=3\cos x+2\sin x$.
4
Note the sign flip
The $-2$ met the $-\sin x$ and became $+2\sin x$. Two minus signs in one term is the classic slip.
EX 2
High-order derivative
$\sin x$, 27th derivativeFind the 27th derivative of $\sin x$.
1
Use the cycle
The derivatives repeat with period $4$: $\sin,\ \cos,\ -\sin,\ -\cos$, then back to $\sin$.
2
Reduce the order
$27=4(6)+3$, so the 27th derivative equals the 3rd.
3
Read off the 3rd
1st: $\cos x$. 2nd: $-\sin x$. 3rd: $-\cos x$.
4
Answer
$f^{(27)}(x)=-\cos x$.
5
Why this works
Only the remainder matters. Computing 27 derivatives by hand would be pointless.
EX 3
Exponential and logarithm together
$e^{x}\ln x$Differentiate $g(x)=e^{x}\ln x$ and state its domain.
1
Recognise a product
Two different functions multiplied, so the product rule applies.
2
Differentiate each factor
$\dfrac{d}{dx}\left[e^{x}\right]=e^{x}$ and $\dfrac{d}{dx}[\ln x]=\dfrac1x$.
3
Assemble
$g'(x)=e^{x}\ln x+e^{x}\cdot\dfrac1x$.
4
Factor
$g'(x)=e^{x}\left(\ln x+\dfrac1x\right)$.
5
State the domain
$\ln x$ requires $x\gt 0$, so both $g$ and $g'$ live on $(0,\infty)$.
EX 4
Change of base
$5^{x}$ and $\log_{2}x$Differentiate $5^{x}$ and $\log_{2}x$.
1
Apply the exponential rule
$\dfrac{d}{dx}\left[a^{x}\right]=a^{x}\ln a$, so $\dfrac{d}{dx}\left[5^{x}\right]=5^{x}\ln 5$.
2
Apply the logarithm rule
$\dfrac{d}{dx}\left[\log_{a}x\right]=\dfrac{1}{x\ln a}$, so $\dfrac{d}{dx}\left[\log_{2}x\right]=\dfrac{1}{x\ln 2}$.
3
Check against the base case
Setting $a=e$ gives $e^{x}\ln e=e^{x}$ and $\dfrac{1}{x\ln e}=\dfrac1x$ — both reduce correctly.
4
Note where $\ln a$ sits
It multiplies for the exponential and divides for the logarithm. Swapping them is a common error.
Watch out
$\dfrac{d}{dx}[\cos x]=-\sin x$
The minus sign is the single most-dropped item in this topic. $\sin$ gains no sign; $\cos$ does.
These are radian derivatives
In degrees $\dfrac{d}{dx}[\sin x]=\dfrac{\pi}{180}\cos x$. Every formula in calculus assumes radians.
$\ln x$ needs $x\gt 0$
Its derivative $\tfrac1x$ is defined for all $x\neq0$, but that is wider than the domain of $\ln$ itself. State the restriction.
The argument must be exactly $x$
$\dfrac{d}{dx}\left[\sin(3x)\right]$ is $3\cos(3x)$, not $\cos(3x)$. Anything other than a bare $x$ inside needs the chain rule.