Derivatives of Sine, Cosine, eˣ, and ln x

The power rule handles powers of $x$, but calculus also needs trig, exponential, and logarithmic functions. Their derivatives turn out to be surprisingly neat: the slope of $\sin x$ is $\cos x$, the slope of $e^x$ is $e^x$ itself, and the slope of $\ln x$ is $\dfrac{1}{x}$. All trig in calculus is in radians.

Key ideas

Radians, not degrees

In Grade 11 you probably measured angles in degrees. Calculus uses radians, where $\pi$ radians $= 180^\circ$, so $\dfrac{\pi}{2} = 90^\circ$, $\dfrac{\pi}{3} = 60^\circ$, and $\dfrac{\pi}{6} = 30^\circ$. The derivative rules below are only true in radians. If you use a calculator on the AP exam, make sure it is in radian mode.

The four new derivatives

Function	Derivative
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$e^x$	$e^x$
$\ln x$	$\dfrac{1}{x}$ (for $x \gt 0$)

They work with the constant multiple, sum, and difference rules from the power rule page, so, for example, $\dfrac{d}{dx}\big[3\sin x - x^2\big] = 3\cos x - 2x$.

Why the slope of sin x is cos x

Look at the slopes of $y = \sin x$. At $x = 0$ the graph climbs with slope $1$; at $x = \tfrac{\pi}{2}$ it’s flat (slope $0$); at $x = \pi$ it falls with slope $-1$. Plot those slopes and you get the graph of $\cos x$.

The slopes of $y = \sin x$ (top) are the values of $y = \cos x$ (bottom). The $x$-axis is in radians.

The algebra behind this uses the limit $\displaystyle\lim_{h \to 0} \frac{\sin h}{h} = 1$, which is only true when $h$ is in radians. That’s why radians matter.

The derivative of $\cos x$ works the same way. Its graph falls just after $x = 0$, which is why its derivative, $-\sin x$, has a minus sign.

Why e is special

Every exponential graph $y = b^x$ has a slope at $x = 0$. The number $e \approx 2.718$ is the base where that slope is exactly $1$. With that choice, the slope of $e^x$ at every point equals its height, so $\dfrac{d}{dx}\big[e^x\big] = e^x$.

The natural logarithm $\ln x$ is the inverse of $e^x$. Its slope is $\dfrac{1}{x}$: steep near $x = 0$ and flattening out as $x$ grows.

Coming soon: tan x and friends

The derivatives of $\tan x$, $\cot x$, $\sec x$, and $\csc x$ come from writing them as quotients of sine and cosine. See the quotient rule. Functions like $\sin(2x)$ or $e^{3x}$ need the chain rule, in the next unit.

Worked examples

Example 1: Sine and cosine

Differentiate $y = 3\sin x - 4\cos x$.

Solution. Use the constant multiple rule on each term. Careful with the sign on the cosine term:

$$\frac{dy}{dx} = 3\cos x - 4(-\sin x) = 3\cos x + 4\sin x$$

Example 2: Exponentials and logs

Find $f'(x)$ for $f(x) = 2e^x - 5\ln x + x^2$.

Solution.

$$f'(x) = 2e^x - \frac{5}{x} + 2x$$

Example 3: A tangent line with trig

Find the equation of the tangent line to $y = \sin x$ at $x = \dfrac{\pi}{3}$.

Solution. Point: $\sin \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$. Slope: $y' = \cos x$, and $\cos \dfrac{\pi}{3} = \dfrac{1}{2}$.

$$y - \frac{\sqrt{3}}{2} = \frac{1}{2}\left(x - \frac{\pi}{3}\right)$$

Point-slope form is the cleanest way to leave this answer.

Example 4: Horizontal tangents on an interval

Find the $x$-values in $[0, 2\pi]$ where $f(x) = x + 2\cos x$ has a horizontal tangent.

Solution. $f'(x) = 1 - 2\sin x$. Set it equal to $0$:

$$1 - 2\sin x = 0 \quad\Rightarrow\quad \sin x = \frac{1}{2}$$

In $[0, 2\pi]$, $\sin x = \tfrac{1}{2}$ at $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$ (the special angles $30^\circ$ and $150^\circ$, in radians).

Common mistakes

Getting the sign of the cosine derivative wrong. $\dfrac{d}{dx}[\cos x] = -\sin x$, but $\dfrac{d}{dx}[\sin x] = +\cos x$. A quick check: $\cos x$ is decreasing just after $x = 0$, so its derivative must be negative there.

Working in degrees. $\dfrac{d}{dx}[\sin x] = \cos x$ is only true in radians. Answers like $x = 30$ instead of $x = \dfrac{\pi}{6}$ lose marks, and a calculator in degree mode gives wrong slopes.

Using the power rule on $e^x$. $\dfrac{d}{dx}\big[e^x\big]$ is $e^x$, not $xe^{x - 1}$. The power rule only applies when the exponent is a constant.

Treating constants like functions. $e^2$, $\ln 5$, and $\sin \tfrac{\pi}{4}$ are numbers, so their derivatives are $0$. Only expressions containing $x$ change.

Thinking $\dfrac{d}{dx}[\sin 2x] = \cos 2x$. When the inside is more than just $x$, you need the chain rule (next unit). For now, stick to $\sin x$, $\cos x$, $e^x$, and $\ln x$ exactly.

Practice

1. (Warm-up) Differentiate each.

• (a) $5\cos x$
• (b) $e^x + 4$
• (c) $\ln x - x$

Solution

(a) $-5\sin x$

(b) $e^x$

(c) $\dfrac{1}{x} - 1$

2. (Warm-up) Let $f(x) = \sin x + \cos x$. Find $f'\left(\dfrac{\pi}{4}\right)$.

Solution

$f'(x) = \cos x - \sin x$, so

$$f'\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0$$

The graph has a horizontal tangent at $x = \dfrac{\pi}{4}$.

3. (Warm-up) Find the slope of $y = 4\ln x$ at $x = 2$.

Solution

$y' = \dfrac{4}{x}$, so the slope at $x = 2$ is $\dfrac{4}{2} = 2$.

4. (Core) Find $\dfrac{dy}{dx}$ for $y = 3e^x - 2\sqrt{x} + \ln\big(x^4\big)$, for $x \gt 0$. (Hint: use a log law first.)

Solution

By the power law of logarithms, $\ln\big(x^4\big) = 4\ln x$. Also $2\sqrt{x} = 2x^{1/2}$.

$$\frac{dy}{dx} = 3e^x - x^{-1/2} + \frac{4}{x} = 3e^x - \frac{1}{\sqrt{x}} + \frac{4}{x}$$

5. (Core) Find the tangent line to $y = e^x$ at $x = 0$, and the tangent line to $y = \ln x$ at $x = 1$.

Solution

For $y = e^x$: the point is $(0, 1)$ and the slope is $e^0 = 1$, so $y = x + 1$.

For $y = \ln x$: the point is $(1, 0)$ and the slope is $\dfrac{1}{1} = 1$, so $y = x - 1$.

(The two graphs are reflections of each other in $y = x$, and so are these two tangent lines.)

6. (Core) Find all $x$ in $[0, 2\pi]$ where $f(x) = 2\sin x + x$ has a horizontal tangent.

Solution

$f'(x) = 2\cos x + 1 = 0$, so $\cos x = -\dfrac{1}{2}$.

In $[0, 2\pi]$: $x = \dfrac{2\pi}{3}$ or $x = \dfrac{4\pi}{3}$.

7. (Core) (Calculator allowed.) Let $f(x) = e^x - 4\cos x$. Find $f'(1)$, correct to three decimal places.

Solution

$f'(x) = e^x + 4\sin x$. In radian mode:

$$f'(1) = e + 4\sin 1 \approx 2.71828 + 3.36588 \approx 6.084$$

8. (Challenge) Find the tangent line to $y = \ln x$ that passes through the origin.

Solution

The tangent line at $x = a$ goes through $(a, \ln a)$ with slope $\dfrac{1}{a}$:

$$y - \ln a = \frac{1}{a}(x - a)$$

For it to pass through $(0, 0)$:

$$0 - \ln a = \frac{1}{a}(0 - a) = -1 \quad\Rightarrow\quad \ln a = 1 \quad\Rightarrow\quad a = e$$

The tangent line is $y - 1 = \dfrac{1}{e}(x - e)$, which simplifies to $y = \dfrac{x}{e}$.

9. (Challenge) Find all $x$ in $[0, 2\pi)$ where the tangent lines to $y = \sin x$ and $y = \cos x$ are parallel.

Solution

Parallel means equal slopes:

$$\cos x = -\sin x \quad\Rightarrow\quad \tan x = -1 \quad (\cos x \ne 0)$$

(If $\cos x = 0$ then $\sin x = \pm 1$, so the equation can’t hold. Dividing by $\cos x$ is safe.)

In $[0, 2\pi)$: $x = \dfrac{3\pi}{4}$ or $x = \dfrac{7\pi}{4}$.