Question

Find dy/dx if x log y + y log x = 5

Answer 1

$\textbf{Given:}$

$\mathsf{x\;logy+y\;logx=5}$

$\textbf{To find:}$

$\mathsf{\dfrac{dy}{dx}}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{x\;logy+y\;logx=5}$

$\textsf{Differentiate with respect to x}$

$\mathsf{x\left(\dfrac{1}{y}\right)\dfrac{dy}{dx}+logy(1)+y\left(\dfrac{1}{x}\right)+logx\left(\dfrac{dy}{dx}\right)=0}$

$\mathsf{\left(\dfrac{x}{y}\right)\dfrac{dy}{dx}+logy+\dfrac{y}{x}+logx\left(\dfrac{dy}{dx}\right)=0}$

$\mathsf{\left(\dfrac{x}{y}\right)\dfrac{dy}{dx}+logx\left(\dfrac{dy}{dx}\right)=-\left(logy+\dfrac{y}{x}\right)}$

$\mathsf{\left(\dfrac{x}{y}+logx\right)\dfrac{dy}{dx}=-\left(logy+\dfrac{y}{x}\right)}$

$\mathsf{\left(\dfrac{x+y\,logx}{y}\right)\dfrac{dy}{dx}=-\left(\dfrac{x\,logy+y}{x}\right)}$

$\mathsf{\dfrac{dy}{dx}=-\left(\dfrac{x\,logy+y}{x+y\,logy}\right){\times}\dfrac{y}{x}}$

$\implies\boxed{\mathsf{\dfrac{dy}{dx}=\dfrac{-y(x\,logy+y)}{x(x+y\,logy)}}}$

$\textbf{Find more:}$

If y=x+√(x²-1), then y-x(dy/dx) = ???

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