If $2 x^{y}+3 y^{x}=20$, then $\frac{d y}{d x}$ at $(2,2)$ is equal to :

Given, $2 x^y+3 y^x=20$ ..........(i) <br/><br/>Let $u=x^y$ <br/><br/>On taking log both sides, we get <br/><br/>$\log u=y \log x$ <br/><br/>On differentiating both sides with respect to $x$, we get <br/><br/>$$ \begin{array}{rlrl} & \frac{1}{u} \frac{d u}{d x} =y \frac{1}{x}+\log x \frac{d y}{d x} \\\\ & \Rightarrow \frac{d u}{d x} =u\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) \\\\ & \Rightarrow \frac{d u}{d x} =x^y\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) .........(ii) \end{array} $$ <br/><br/>Also, let $v=y^x$ <br/><br/>On taking log both sides, we get <br/><br/>$\log v=x \log y$ <br/><br/>On differentiating both sides, we get <br/><br/>$$ \begin{array}{rlrl} & \frac{1}{v} \frac{d v}{d x} =x \frac{1}{y} \frac{d y}{d x}+\log y \cdot 1 \\\\ &\Rightarrow \frac{d v}{d x} =v\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) \\\\ &\Rightarrow \frac{d v}{d x} =y^x\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) ..........(iii) \end{array} $$ <br/><br/>Now, from Equation (i), $2 u+3 v=20$ <br/><br/>$$ \begin{aligned} & \Rightarrow 2 \frac{d u}{d x}+3 \frac{d v}{d x}=0 \\\\ & \Rightarrow 2 x^y\left(\frac{y}{x}+\log x \frac{d y}{d x}\right)+3 y^x\left(\frac{x}{y} \frac{d y}{d x}+\log y\right)=0 \text { [Using Eqs. (ii) and (iii)] } \end{aligned} $$ <br/><br/>On putting $x=2$ and $y=2$, we get <br/><br/>$$ \begin{aligned} & 2(4)\left(1+\log 2 \frac{d y}{d x}\right)+3(4)\left(\frac{d y}{d x}+\log 2\right)=0 \\\\ & \Rightarrow \frac{d y}{d x}(8 \log 2+12)+(8+12 \log 2)=0 \\\\ & \Rightarrow \frac{d y}{d x}=-\left(\frac{2+3 \log 2}{3+2 \log 2}\right) \\\\ & \Rightarrow \frac{d y}{d x}=-\left(\frac{2+\log 8}{3+\log 4}\right) \end{aligned} $$