y = x^2/2 + cosax + be^(bx) + log(c+x) + d  ধরলে dy/dx = 

প্রশ্নানুসারে, \(y = \frac{x^2}{2} + \cos(ax) + be^{bx} + \log(c+x) + d\)

\(\frac{dy}{dx}\) নির্ণয় করতে হবে।

আমরা জানি,

1. \(\frac{d}{dx} x^n = nx^{n-1}\)
2. \(\frac{d}{dx} \cos(ax) = -a\sin(ax)\)
3. \(\frac{d}{dx} e^{bx} = be^{bx}\)
4. \(\frac{d}{dx} \log(x) = \frac{1}{x}\)
5. \(\frac{d}{dx} \) ধ্রুবক \(= 0\)

অতএব,

\(\frac{dy}{dx} = \frac{d}{dx} \left(\frac{x^2}{2}\right) + \frac{d}{dx} \left(\cos(ax)\right) + \frac{d}{dx} \left(be^{bx}\right) + \frac{d}{dx} \left(\log(c+x)\right) + \frac{d}{dx} (d)\)

\(= \frac{1}{2} \cdot 2x + (-a\sin(ax)) + b \cdot be^{bx} + \frac{1}{c+x} + 0\)

\(= x - a\sin(ax) + b^2e^{bx} + \frac{1}{c+x}\)

সুতরাং, \(\frac{dy}{dx} = x - a\sin(ax) + b^2e^{bx} + \frac{1}{c+x}\) 🥳