If 2 x y + 3 y x = 20 , then d y d x at 2 , 2 is equal to:

Given,

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

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

$2x^y\left(y^{\text{'}}\ln x+\frac{y}{x}\right)+3y^x\left(\ln y+\frac{x}{y}·y^{\text{'}}\right)=0$ $\left\{\text{here }\ln ={\log }_e\right\}$

Putting $x=2 \& y=2,$ we get

$\Rightarrow 8\left(\ln 2·y^{\text{'}}+1\right)+12\left(\ln y+y^{\text{'}}\right)=0$

$\Rightarrow 8\left(\ln 2·y^{\text{'}}+1\right)+12\left(\ln y+y^{\text{'}}\right)=0$

$\Rightarrow y^{\text{'}}=-\left(\frac{3\ln 2+2}{2\ln 2+3}\right)$

$\Rightarrow y^{\text{'}}=-\left(\frac{2+\ln 8}{3+\ln 4}\right)$