Question:
If \(y = e^{x^2}\sin 2x\), then what is \(\frac{dy}{dx}\) at \(x = \pi\) equal to?
यदि \(y = e^{x^2}\sin 2x\), तो \(x = \pi\) पर \(\frac{dy}{dx}\) का मान किसके बराबर है?
Options:
- \((1+\pi)e^{\pi^2}\)
- \(2\pi e^{\pi^2}\)
- \(2e^{\pi^2}\)
- \(e^{\pi^2}\)
- \((1+\pi)e^{\pi^2}\)
- \(2\pi e^{\pi^2}\)
- \(2e^{\pi^2}\)
- \(e^{\pi^2}\)
Correct Answer: c)
Explanation:
Using the product rule for differentiation:
\[\frac{dy}{dx} = \frac{d}{dx}(e^{x^2}) \cdot \sin 2x + e^{x^2} \cdot \frac{d}{dx}(\sin 2x) = 2x e^{x^2}\sin 2x + 2 e^{x^2}\cos 2x\]
At \(x = \pi\):
\(\sin(2\pi) = 0\) and \(\cos(2\pi) = 1\).
\[\left.\frac{dy}{dx}\right|_{x = \pi} = 2\pi e^{\pi^2}(0) + 2 e^{\pi^2}(1) = 2e^{\pi^2}\]
Hence, the correct option is (c).
गुणनफल नियम से अवकलन करने पर:
\[\frac{dy}{dx} = 2x e^{x^2}\sin 2x + 2e^{x^2}\cos 2x\]
\(x = \pi\) रखने पर:
चूँकि \(\sin(2\pi) = 0\) तथा \(\cos(2\pi) = 1\) है:
\[\left.\frac{dy}{dx}\right|_{x = \pi} = 0 + 2e^{\pi^2}(1) = 2e^{\pi^2}\]
अतः सही विकल्प (c) है।