If 2 x ^ y + 3 y ^ x = 20, then (dy)/(dx) at (2, 2) is equal to:

Given,

$2{x}^{y} + 3{y}^{x} = 20$

Now differentiating both sides w.r.t $x,$ we get

$2{x}^{y}\left({y}^{'}\ln x + \dfrac{y}{x}\right) + 3{y}^{x}\left(\ln y + \dfrac{x}{y} \cdot {y}^{'}\right) = 0$ \left\{\text{here}ln = \log_{e}\right\}$</p><p>Putting$x = 2 \text{ and } y = 2,$we get</p><p>$⇒ 8\left(\ln 2 \cdot {y}^{'} + 1\right) + 12\left(\ln y + {y}^{'}\right) = 0$</p><p>$⇒ 8\left(\ln 2 \cdot {y}^{'} + 1\right) + 12\left(\ln y + {y}^{'}\right) = 0$</p><p>$⇒ {y}^{'} = - \left(\dfrac{3\ln 2 + 2}{2\ln 2 + 3}\right)$</p><p>$⇒ {y}^{'} = - \left(\dfrac{2 + \ln 8}{3 + \ln 4}\right)