Pure · Differentiation · Year 2

Differentiating Trig Functions (Reciprocal & Inverse)

Standard derivatives for sec, cosec, cot, and the three inverse trig functions — the last set of standard results before the chain rule ties everything together.

9 min read 3 worked examples

Differentiating the reciprocal trig functions

Recall the three reciprocal trig functions: $\sec x=\dfrac{1}{\cos x}$, $\csc x=\dfrac{1}{\sin x}$, and $\cot x=\dfrac{1}{\tan x}$. Their derivatives are standard results — a full derivation using the quotient rule appears in a later lesson, but the results themselves are needed right away:

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

A pattern worth noticing

Each "co-" function ($\csc$, $\cot$, and $\cos$ itself) differentiates to something with a negative sign, while $\sec$, $\tan$, and $\sin$ do not. Keeping track of which functions start with "co-" is a quick way to remember which derivatives carry the minus sign.

Differentiating the inverse trig functions

The inverse trig functions $\arcsin x$, $\arccos x$, and $\arctan x$ (also written $\sin^{-1}x$, $\cos^{-1}x$, $\tan^{-1}x$) also have standard derivatives — proved properly using implicit differentiation in a later lesson:

$\dfrac{d}{dx}(\arcsin x)=\dfrac{1}{\sqrt{1-x^2}} \qquad \dfrac{d}{dx}(\arccos x)=-\dfrac{1}{\sqrt{1-x^2}} \qquad \dfrac{d}{dx}(\arctan x)=\dfrac{1}{1+x^2}$

Domains matter

$\arcsin x$ and $\arccos x$ are only defined for $-1\le x\le1$, and their derivatives are undefined at the endpoints $x=\pm1$ (the curve becomes vertical there). $\arctan x$, by contrast, is defined — and differentiable — for every real $x$, since $1+x^2$ is never zero.

Worked example 1

Reciprocal trig derivatives

Differentiate $y=3\sec x-2\cot x$.

MethodApply each standard result

$\dfrac{d}{dx}(3\sec x)=3\sec x\tan x$.

Standard result, scaled by 3.

$\dfrac{d}{dx}(-2\cot x)=-2(-\csc^2x)=2\csc^2x$.

Standard result, scaled by $-2$ — two negatives make a positive.

$\dfrac{dy}{dx}=3\sec x\tan x+2\csc^2x$.

Add the two terms.

▪ solved

Worked example 2

Gradient of an inverse trig curve

Find the gradient of $y=\arcsin x$ at the point where $x=\dfrac12$.

MethodDifferentiate, then substitute

$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}$.

Standard result for $\arcsin x$.

At $x=\dfrac12$: $1-x^2=1-\dfrac14=\dfrac34$.

Substitute $x=\dfrac12$ into $1-x^2$.

$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{3/4}}=\dfrac{1}{\frac{\sqrt3}{2}}=\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}$.

Simplify the surd, then rationalise the denominator.

▪ solved

Worked example 3

Combining reciprocal and inverse trig · Challenge

Find $\dfrac{dy}{dx}$ at $x=0$ for the curve $y=\arccos x-\sec x$.

MethodDifferentiate term by term, then substitute x=0

$\dfrac{dy}{dx}=-\dfrac{1}{\sqrt{1-x^2}}-\sec x\tan x$.

Standard results for $\arccos x$ and $\sec x$.

At $x=0$: $\sqrt{1-0^2}=1$, so the first term is $-1$.

Substitute $x=0$ into the arccos term.

At $x=0$: $\sec0=1$, $\tan0=0$, so the second term is $-(1)(0)=0$.

Substitute $x=0$ into the sec term.

$\dfrac{dy}{dx}=-1-0=-1$ at $x=0$.

Add the two evaluated terms.

▪ solved

Look deeper

$x=0$ sits comfortably inside the domain of $\arccos x$ ($-1\le x\le1$) and away from any point where $\sec x$ is undefined, so no domain issues arise here — always worth a quick check before evaluating a derivative built from functions with restricted domains.

Common mistake

Mixing up $\arcsin x$ (notation $\sin^{-1}x$) with $\dfrac{1}{\sin x}=\csc x$ — the "$-1$" in $\sin^{-1}x$ means inverse function, not reciprocal, and the two differentiate completely differently. Also common: dropping the minus sign in the derivatives of $\csc x$, $\cot x$, and $\arccos x$.

Check yourself

Differentiate $y=\csc x+\arctan x$.

Method

$\dfrac{d}{dx}(\csc x)=-\csc x\cot x$.

Standard result.

$\dfrac{d}{dx}(\arctan x)=\dfrac{1}{1+x^2}$.

Standard result.

$\dfrac{dy}{dx}=-\csc x\cot x+\dfrac{1}{1+x^2}$.

Add the two terms.

▪ solved

← Previous

Differentiating Exponentials and Logarithms

The Chain Rule