यदि y = frac{alpha x + beta}{gamma x + delta} | vedclass

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(B) दिया गया है $y = \frac{\alpha x + \beta}{\gamma x + \delta}$.प्रथम अवकलज: $y_1 = \frac{\alpha(\gamma x + \delta) - \gamma(\alpha x + \beta)}{(\gam

Vedclass · Accepted Answer

$3 y_2^2$

Vedclass · Answer

(B) दिया गया है $y = \frac{\alpha x + \beta}{\gamma x + \delta}$.प्रथम अवकलज: $y_1 = \frac{\alpha(\gamma x + \delta) - \gamma(\alpha x + \beta)}{(\gamma x + \delta)^2} = \frac{\alpha \delta - \gamma \beta}{(\gamma x + \delta)^2}$.द्वितीय अवकलज: $y_2 = \frac{d}{dx} [(\alpha \delta - \gamma \beta)(\gamma x + \delta)^{-2}] = -2(\alpha \delta - \gamma \beta)(\gamma x + \delta)^{-3} \cdot \gamma = \frac{-2 \gamma(\alpha \delta - \gamma \beta)}{(\gamma x + \delta)^3}$.तृतीय अवकलज: $y_3 = \frac{d}{dx} [-2 \gamma(\alpha \delta - \gamma \beta)(\gamma x + \delta)^{-3}] = (-2 \gamma)(\alpha \delta - \gamma \beta) \cdot (-3)(\gamma x + \delta)^{-4} \cdot \gamma = \frac{6 \gamma^2(\alpha \delta - \gamma \beta)}{(\gamma x + \delta)^4}$.अब,$2 y_1 y_3 = 2 \cdot \left( \frac{\alpha \delta - \gamma \beta}{(\gamma x + \delta)^2} \right) \cdot \left( \frac{6 \gamma^2(\alpha \delta - \gamma \beta)}{(\gamma x + \delta)^4} \right) = \frac{12 \gamma^2(\alpha \delta - \gamma \beta)^2}{(\gamma x + \delta)^6}$.साथ ही,$3 y_2^2 = 3 \cdot \left( \frac{-2 \gamma(\alpha \delta - \gamma \beta)}{(\gamma x + \delta)^3} \right)^2 = 3 \cdot \frac{4 \gamma^2(\alpha \delta - \gamma \beta)^2}{(\gamma x + \delta)^6} = \frac{12 \gamma^2(\alpha \delta - \gamma \beta)^2}{(\gamma x + \delta)^6}$.अतः,$2 y_1 y_3 = 3 y_2^2$.

यदि $y = \frac{\alpha x + \beta}{\gamma x + \delta}$ है,तो $2 y_1 y_3 =$

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