Question:
What is the derivative of \(\sec^2(\tan^{-1} x)\) with respect to \(x\)?
\(x\) के सापेक्ष \(\sec^2(\tan^{-1} x)\) का अवकलज (derivative) क्या है?
Options:
- \(2x\)
- \(x^2 + 1\)
- \(x + 1\)
- \(x^2\)
- \(2x\)
- \(x^2 + 1\)
- \(x + 1\)
- \(x^2\)
Correct Answer: a)
Explanation:
Using the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\):
Let \(\theta = \tan^{-1} x\), then \(\tan \theta = x\).
\[\sec^2(\tan^{-1} x) = 1 + [\tan(\tan^{-1} x)]^2 = 1 + x^2\]
Differentiating with respect to \(x\):
\[\frac{d}{dx}(1 + x^2) = 2x\]
Hence, the correct option is (a).
त्रिकोणमितीय सर्वसमिका \(\sec^2 \theta = 1 + \tan^2 \theta\) का उपयोग करने पर:
\(\theta = \tan^{-1} x\) रखने पर, \(\tan \theta = x\) प्राप्त होता है।
\[\sec^2(\tan^{-1} x) = 1 + x^2\]
\(x\) के सापेक्ष अवकलन करने पर:
\[\frac{d}{dx}(1 + x^2) = 2x\]
अतः सही विकल्प (a) है।