Q.State whether True or False: Trigonometric and inverse-trigonometric functions are differentiable in their respective domains.

Check the two families against the definition of differentiability on their whole domains.

• Trigonometric functions ($\sin x,\cos x,\tan x,\dots$) are differentiable at every point where they are defined — no problem here.
• Inverse-trigonometric functions are the issue. Their domains are closed: $\sin^{-1}x$ and $\cos^{-1}x$ have domain $[-1,1]$, and $\sec^{-1}x,\csc^{-1}x$ have domain $|x|\ge 1$. At the endpoints the tangent is vertical: e.g. $\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}\to\infty$ as $x\to\pm1$. So $\sin^{-1}x$ is not differentiable at $x=\pm1$, which are points of its …

What the statement claims

It asserts that both the trigonometric functions and their inverses are differentiable everywhere in their respective domains. To judge it, we test each family at every point of its domain — including any boundary points that belong to the domain.

The trigonometric functions are fine

For $\sin x$, $\cos x$, $\tan x$, $\cot x$, $\sec x$, $\csc x$, the derivative exists at every point where the function is defined:

The places where, say, $\tan x$ misbehaves ($x=\tfrac{\pi}{2}$) are not in its domain, so they don't count against it. So for the ordinary trig functions the claim is true.

The inverse-trigonometric functions break the claim

The trouble is the endpoints of the closed domains:

• $\sin^{-1}x$ and $\cos^{-1}x$ have domain $[-1,1]$.
• $\sec^{-1}x$ and $\csc^{-1}x$ have domain $(-\infty,-1]\cup[1,\infty)$.

Look at the derivative of $\sin^{-1}x$:

As $x\to\pm1$, the denominator $\sqrt{1-x^2}\to 0$, so the derivative $\to\infty$. Geometrically the graph of $\sin^{-1}x$ has a vertical tangent at $x=\pm1$, so no finite derivative exists there. Yet $x=\pm1$ are points of the domain $[-1,1]$. The same happens for $\cos^{-1}x$ at $x=\pm1$, and for $\sec^{-1}x,\csc^{-1}x$ at $x=\pm1$. …

Method: Checking Differentiability at the Endpoints of a Closed Domain

This method solves "is this function differentiable throughout its domain" questions, especially where the domain is a closed interval — as it is for several inverse trigonometric functions.

Steps

Step 1: Identify the domain and note whether it is open or closed

Differentiability requires a well-defined two-sided limit for the difference quotient, so a domain that includes its boundary points deserves special scrutiny — behaviour at an included endpoint is not automatically the same as in the interior.

Step 2: Write down the function's standard derivative formula

For example, $\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}$, valid where the formula makes sense.

Step 3: Examine the derivative's behaviour as $x$ approaches each domain endpoint …

Common Mistakes

Mistake 1: Treating "domain" as an open interval by default

Students often check differentiability of $\sin^{-1}x$ only on the open interval $(-1,1)$, silently excluding the endpoints, and conclude the statement is True. Why it's wrong: the actual domain of $\sin^{-1}x$ is the CLOSED interval $[-1,1]$, and $x=\pm1$ genuinely belong to that domain — the statement claims differentiability at EVERY domain point, including those endpoints, where $\frac{1}{\sqrt{1-x^2}}$ blows up. Correct approach: always check the function's full domain as officially defined, closed endpoints included, not an informal "nice" open version of it.

Mistake 2: Assuming inverse trig functions inherit their originals' good behavior …