What is the slope of the tangent line of $(y/x)e^(x/y)= C$ , where C is an arbitrary constant, at $(1,2)$ ?

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Answer 1 · 2016-01-12T09:50:03

slope $= 2$

Start with the given $(y/x)e^(x/y) = C$ at $(1, 2)$

obtain the derivative of both sides of the equation

$(y/x)e^(x/y)*((y*1-x*y')/y^2)+e^(x/y)*((xy'-y*1)/x^2)=0$

After dividing both sides by $e^(x/y)$

$(y/x)*((y*1-x*y')/y^2)+((xy'-y*1)/x^2)=0$

$-y'/y+y'/x=y/x^2-1/x$

$y'(1/x-1/y)=y/x^2-1/x$

$y' = (y/x^2-1/x)/(1/x-1/y)$

substitute now the values $x=1$ and $y=2$

$y' = (2/1^2-1/1)/(1/1-1/2)=(2-1)/(1-1/2)=2$

What is the slope of the tangent line of (y/x)e^(x/y)= C (y/x)e^(x/y)= C , where C is an arbitrary constant, at (1,2)(1,2)?