Differentiating tan x and the Reciprocal Ratios

tan x is a quotient

$\tan x = \frac{{\sin x}}{{\cos x}}$, defined wherever $\cos x \ne 0$, which is every x except $\frac{\pi}{2} + n\pi$ for whole numbers n. Both parts have known derivatives, so the quotient rule applies, with x in radians throughout.

The numerator is u = sin x, so u′ = cos x. The denominator is v = cos x, so v′ = −sin x.

The quotient rule on sin x over cos x

The quotient rule gives $\frac{{u'v - uv'}}{v^{2}} = \frac{{\cos x \cdot \cos x - \sin x \cdot \left(-\sin x\right)}}{\cos^{2}x}$.

The second term is sin x times −sin x, subtracted, so the two minus signs make a plus: the numerator is $\cos^{2}x + \sin^{2}x$.

The numerator is 1

By Pythagoras, $\cos^{2}x + \sin^{2}x = 1$ for every x. So the derivative of tan x is $\frac{1}{{\cos^{2}x}}$, and $\frac{1}{{\cos x}}$ is sec x, so this is $\sec^{2}x$.

Since $\cos^{2}x$ is at most 1, $\sec^{2}x$ is at least 1. So tan x rises on every branch of its graph, with gradient never less than 1, and exactly 1 where $\cos^{2}x = 1$, at $x = n\pi$.

Check at three points. At x = 0, $\sec^{2}0 = 1$. At $x = \frac{\pi}{4}$, $\cos^{2}x = \frac{1}{2}$, so $\sec^{2}x = 2$; the chord from $\frac{\pi}{4}$ to $\frac{\pi}{4} + 0.001$ has gradient 2.0020. At x = 1, $\sec^{2}1 = 3.4255$, and a chord with a step of 0.000001 gives 3.4255.

The gold curve y = tan x, for x from −1.5 to 1.5 radians. The dashed line is its tangent at the origin, with gradient $\sec^{2}0 = 1$. The gold line is its tangent at $\left(\frac{\pi}{4} , 1\right)$, with gradient $\sec^{2}\left(\frac{\pi}{4}\right) = 2$.

sec x

$\sec x = \frac{1}{{\cos x}} = \left(\cos x\right)^{-1}$, defined where $\cos x \ne 0$. The chain rule applies, with inner function cos x and outer function $u^{-1}$. The outer derivative is $-u^{-2}$, and the inner derivative is −sin x.

So the derivative is $-\left(\cos x\right)^{-2} \times \left(-\sin x\right) = \frac{{\sin x}}{{\cos^{2}x}}$. Split it as $\left(\frac{1}{{\cos x}}\right) \times \left(\frac{{\sin x}}{{\cos x}}\right)$, and that is sec x tan x.

Check at $x = \frac{\pi}{3}$, where sec x = 2 and $\tan x = \sqrt{3}$: the derivative is $2\sqrt{3} = 3.4641$, and a chord with a step of 0.000001 gives 3.4641. At x = 0.5 the derivative is 0.6225, and the chord gives 0.6225.

cosec x and cot x

$\cosec x = \frac{1}{{\sin x}} = \left(\sin x\right)^{-1}$, defined where $\sin x \ne 0$. The same chain-rule move gives $-\left(\sin x\right)^{-2} \times \cos x = - \frac{{\cos x}}{{\sin^{2}x}}$, which splits as −cosec x cot x. At $x = \frac{\pi}{6}$, cosec x = 2 and $\cot x = \sqrt{3}$, so the derivative is $-2\sqrt{3} = -3.4641$. At x = 1 it is −0.7631. Chords give the same two values.

$\cot x = \frac{{\cos x}}{{\sin x}}$, so the quotient rule applies, with u' = −sin x and v' = cos x. The numerator is $-\sin x \cdot \sin x - \cos x \cdot \cos x = -\left(\sin^{2}x + \cos^{2}x\right) = -1$, so the derivative is $- \frac{1}{{\sin^{2}x}} = -\cosec^{2}x$. At $x = \frac{\pi}{4}$ that is −2, and at x = 1 it is −1.4123. Chords give the same two values.

cot x is also $\frac{1}{{\tan x}}$. The chain rule on $\left(\tan x\right)^{-1}$ gives $- \frac{\sec^{2}x}{\tan^{2}x}$, and since $\frac{\sec^{2}x}{\tan^{2}x} = \left(\frac{1}{\cos^{2}x}\right) \times \left(\frac{\cos^{2}x}{\sin^{2}x}\right) = \frac{1}{\sin^{2}x}$, that is $-\cosec^{2}x$ again.

The four results together

tan x differentiates to $\sec^{2}x$, and sec x to sec x tan x. The co-functions follow the same pattern with cot and cosec in place of tan and sec, and a minus sign: cot x differentiates to $-\cosec^{2}x$, and cosec x to −cosec x cot x.

The minus signs match the graphs. On $0 < x < \frac{\pi}{2}$, cot x and cosec x are both falling, while tan x and sec x are both rising.

A coefficient on the angle

The chain rule brings out the inner derivative, as before. For y = tan 3x, the inner function is 3x and the outer function is tan, so $\frac{dy}{dx} = 3 \sec^{2}3x$, which is 3 at x = 0. For y = sec 2x, $\frac{dy}{dx} = 2 \sec 2x \tan 2x$, which is 1.6578 at x = 0.3.

The usual mistakes

Giving sec x tan x as the derivative of tan x. That is the derivative of sec x; tan x gives $\sec^{2}x$.

Differentiating sin x / cos x term by term, to get $\frac{{\cos x}}{{-\sin x}}$. The quotient rule is needed, and it gives $\sec^{2}x$.

Dropping the minus from cot or cosec. Both functions fall on $0 < x < \frac{\pi}{2}$, so both derivatives are negative there.

Adding a minus to sec x. The −1 from the power and the minus in −sin x multiply to +1, so sec x differentiates to plus sec x tan x.

A lighthouse beam

In the application below, a lit spot is $300 \tan \theta$ meters along a wall. Its derivative, $300 \sec^{2}\theta$, is in meters per radian, and the chain rule turns it into meters per second.