Question

If \[ y=\sec^{-1}\left(\frac{1+x^2}{2x}\right) \quad \text{and} \quad x>1, \] then \[ \frac{dy}{dx}= \]

• A. \(\dfrac{1}{1+x^2}\)
• B. \(\dfrac{2}{1+x^2}\)
• C. \(-\dfrac{1}{1+x^2}\)
• D. \(-\dfrac{2}{1+x^2}\)

Hint:

Whenever you see expressions of the form \[ \frac{1+x^2}{2x}, \] try converting them into tangent half-angle identities. This frequently simplifies inverse trigonometric differentiation problems.

Answer 1

The Correct Option is B

Step 1: Spot a known identity.
We have $y=\sec^{-1}\left(\dfrac{1+x^2}{2x}\right)$. The trick is to write the inside as a standard form so the inverse simplifies.
Step 2: Use the double-angle form.
Recall that $\sec(2\tan^{-1}x)=\dfrac{1+x^2}{2x}$, since $\cos(2\tan^{-1}x)=\dfrac{1-x^2}{1+x^2}$ and its companion gives $\dfrac{2x}{1+x^2}$ for cosine, hence $\dfrac{1+x^2}{2x}$ for secant.
Step 3: Simplify $y$.
For $x>1$ this lets us write $y=2\tan^{-1}x$. The messy inverse secant has turned into a simple inverse tangent.
Step 4: Recall the derivative of $\tan^{-1}x$.
$\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}$.
Step 5: Differentiate.
\[ \frac{dy}{dx}=2\cdot\frac{1}{1+x^2}=\frac{2}{1+x^2}. \]
Step 6: State the answer.
The derivative is \[ \boxed{\frac{2}{1+x^2}}. \]