Q.Find $\frac{dy}{dx}$ in the following: $y = \cos^{-1} \left(\frac{2x}{1+x^2}\right), -1 < x < 1$

Step 1: Let $x = \tan\theta$. Since $-1 < x < 1$, $\theta \in \left(-\tfrac{\pi}{4},\tfrac{\pi}{4}\right)$. Then

$\frac{2x}{1+x^2} = \sin 2\theta, \qquad y = \cos^{-1}(\sin 2\theta) = \cos^{-1}\!\left[\cos\!\left(\frac{\pi}{2}-2\theta\right)\right].$ …

Recognising the Identity

The argument $\dfrac{2x}{1+x^2}$ is the double-angle formula $\sin 2\theta = \dfrac{2\tan\theta}{1+\tan^2\theta}$ in terms of $x = \tan\theta$. Because the outer function here is $\cos^{-1}$, the sine must first be converted to a cosine before the inverse function can cancel it.

Step-by-Step Solution

1. Substitute $x = \tan\theta$.

Since $-1 < x < 1$, let $\theta = \tan^{-1}x \in \left(-\tfrac{\pi}{4}, \tfrac{\pi}{4}\right)$. Then:

2. Rewrite the sine as a cosine.

Using $\sin\phi = \cos\!\left(\tfrac{\pi}{2}-\phi\right)$ with $\phi = 2\theta$:

3. Check the range.

Since $\theta \in \left(-\tfrac{\pi}{4}, \tfrac{\pi}{4}\right)$, $2\theta \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$, so:

the principal range of $\cos^{-1}$. So the inverse cosine cancels the cosine directly:

4. Differentiate. …

Method: Converting Between Inverse-Cosine and Inverse-Sine Forms via the Complementary Identity

This is the mirror image of the sine-to-cosine conversion: here the double-angle substitution naturally produces a sine ($\sin 2\theta$), but the outer inverse function is $\cos^{-1}$, so a complementary-angle conversion is needed before cancelling.

Steps

Step 1: Substitute and identify the mismatch

With $x=\tan\theta$, $\frac{2x}{1+x^2}=\sin 2\theta$, so the expression is $y=\cos^{-1}(\sin 2\theta)$ — a sine trapped inside an inverse cosine.

Step 2: Convert the sine to a cosine using the complementary identity

With $\phi=2\theta$: $\sin 2\theta=\cos\!\left(\frac{\pi}{2}-2\theta\right)$, giving $y=\cos^{-1}\!\left[\cos\!\left(\frac{\pi}{2}-2\theta\right)\right]$.

Step 3: Check the resulting angle against $\cos^{-1}$'s range $[0,\pi]$

$\cos^{-1}(\cos\alpha)=\alpha$ only when $\alpha\in[0,\pi]$ — a wider, differently-positioned interval than $\sin^{-1}$'s, so re-derive it fresh rather than reusing a check from a sine problem. …

Common Mistakes

Mistake 1: Forgetting cosine's inverse needs a cosine, not a sine, before cancelling

Since $\frac{2x}{1+x^2}$ substitutes directly to $\sin 2\theta$, it's tempting to plug that straight into $\cos^{-1}$ and simplify without converting. Why it's wrong: $\cos^{-1}$ only cancels a $\cos$, so a $\sin 2\theta$ trapped inside it must first be rewritten as $\cos\left(\frac{\pi}{2}-2\theta\right)$ via the complementary identity. Correct approach: always check whether the inner trig function matches the outer inverse function's own type; convert with the complementary identity if it doesn't.

Mistake 2: Reusing a range check from a different (sine) problem …