Differentiating tan x and the Reciprocal Ratios

tan x is a quotient

$\tan x=\frac{\sin x}{\cos x}$tan x = (sin x)/(cos x), defined wherever $\cos x\ne 0$cos x ≠ 0, which is every x except $\frac{\pi }{2}+n\pi$π/2 + nπ 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 $\frac{u^{\prime }v-u{v}^{\prime }}{v^2}=\frac{\cos x\cdot \cos x-\sin x\cdot \left(-\sin x\right)}{{\cos }^{2}x}$(u'v − uv')/v² = (cos x · cos x − sin x · (−sin x))/cos²x.

The second term is sin x times −sin x, subtracted, so the two minus signs make a plus: the numerator is ${\cos }^{2}x+{\sin }^{2}x$cos²x + sin²x.

The numerator is 1

By Pythagoras, ${\cos }^{2}x+{\sin }^{2}x=1$cos²x + sin²x = 1 for every x. So the derivative of tan x is $\frac{1}{{\cos }^{2}x}$1/(cos²x), and $\frac{1}{\cos x}$1/(cos x) is sec x, so this is ${\sec }^{2}x$sec²x.

Since ${\cos }^{2}x$cos²x is at most 1, ${\sec }^{2}x$sec²x is at least 1. So tan x rises on every branch of its graph, with gradient never less than 1, and exactly 1 where ${\cos }^{2}x=1$cos²x = 1, at $x=n\pi$x = nπ.

Check at three points. At x = 0, ${\sec }^{2}0=1$sec²0 = 1. At $x=\frac{\pi }{4}$x = π/4, ${\cos }^{2}x=\frac{1}{2}$cos²x = 1/2, so ${\sec }^{2}x=2$sec²x = 2; the chord from $\frac{\pi }{4}$π/4 to $\frac{\pi }{4}+0.001$π/4 + 0.001 has gradient 2.0020. At x = 1, ${\sec }^{2}1=3.4255$sec²1 = 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 ${\sec }^{2}0=1$sec²0 = 1. The gold line is its tangent at $\left(\frac{\pi }{4},1\right)$(π/4, 1), with gradient ${\sec }^2\left(\frac{\pi }{4}\right)=2$sec²(π/4) = 2.

sec x

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

So the derivative is $-{\left(\cos x\right)}^{-2}\times \left(-\sin x\right)=\frac{\sin x}{{\cos }^{2}x}$−(cos x)⁻² × (−sin x) = (sin x)/(cos²x). Split it as $\left(\frac{1}{\cos x}\right)\times \left(\frac{\sin x}{\cos x}\right)$(1/(cos x)) × ((sin x)/(cos x)), and that is sec x tan x.

Check at $x=\frac{\pi }{3}$x = π/3, where sec x = 2 and $\tan x=\sqrt{3}$tan x = √3: the derivative is $2\sqrt{3}=3.4641$2√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

$\mathrm{cosec}x=\frac{1}{\sin x}={\left(\sin x\right)}^{-1}$cosec x = 1/(sin x) = (sin x)⁻¹, defined where $\sin x\ne 0$sin x ≠ 0. The same chain-rule move gives $-{\left(\sin x\right)}^{-2}\times \cos x=-\frac{\cos x}{{\sin }^{2}x}$−(sin x)⁻² × cos x = −(cos x)/(sin²x), which splits as −cosec x cot x. At $x=\frac{\pi }{6}$x = π/6, cosec x = 2 and $\cot x=\sqrt{3}$cot x = √3, so the derivative is $-2\sqrt{3}=-3.4641$−2√3 = −3.4641. At x = 1 it is −0.7631. Chords give the same two values.

$\cot x=\frac{\cos x}{\sin x}$cot x = (cos x)/(sin x), so the quotient rule applies, with u' = −sin x and v' = cos x. The numerator is $-\sin x\cdot \sin x-\cos x\cdot \cos x=-\left({\sin }^{2}x+{\cos }^{2}x\right)=-1$−sin x · sin x − cos x · cos x = −(sin²x + cos²x) = −1, so the derivative is $-\frac{1}{{\sin }^{2}x}=-{\mathrm{cosec}}^{2}x$−1/(sin²x) = −cosec²x. At $x=\frac{\pi }{4}$x = π/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}$1/(tan x). The chain rule on ${\left(\tan x\right)}^{-1}$(tan x)⁻¹ gives $-\frac{{\sec }^{2}x}{{\tan }^{2}x}$−sec²x/tan²x, and since $\frac{{\sec }^{2}x}{{\tan }^{2}x}=\left(\frac{1}{{\cos }^{2}x}\right)\times \left(\frac{{\cos }^{2}x}{{\sin }^{2}x}\right)=\frac{1}{{\sin }^{2}x}$sec²x/tan²x = (1/cos²x) × (cos²x/sin²x) = 1/sin²x, that is $-{\mathrm{cosec}}^{2}x$−cosec²x again.

The four results together

tan x differentiates to ${\sec }^{2}x$sec²x, 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 $-{\mathrm{cosec}}^{2}x$−cosec²x, and cosec x to −cosec x cot x.

The minus signs match the graphs. On $0<x<\frac{\pi }{2}$0 < x < π/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 $\frac{dy}{dx}=3{\sec }^{2}3x$dy/dx = 3 sec²3x, which is 3 at x = 0. For y = sec 2x, $\frac{dy}{dx}=2\sec 2x\tan 2x$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 ${\sec }^{2}x$sec²x.

Differentiating sin x / cos x term by term, to get $\frac{\cos x}{-\sin x}$(cos x)/(−sin x). The quotient rule is needed, and it gives ${\sec }^{2}x$sec²x.

Dropping the minus from cot or cosec. Both functions fall on $0<x<\frac{\pi }{2}$0 < x < π/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$300 tan θ meters along a wall. Its derivative, $300{\sec }^2\theta$300 sec²θ, is in meters per radian, and the chain rule turns it into meters per second.