Find the slope of the tangent to the curve x^y = y^x at the point (3, 6)

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find the slope of the tangent to the curve x^y = y^x at the point (3, 6). The slope is given as the value for dy/dx at the particular point.

x^y = y^x

take the log of both the sides

=> y*ln x = x*ln y

differentiate both the side

=> (y/x) + ln x *(dy/dx) = (x/y)*(dy/dx) + ln y

=> ln x * (dy/dx) - x*(dy/dx)/y = ln y - (y/x)

=> (dy/dx)*(ln x - (x/y)) = ln y - (y/x)

=> dy/dx = (ln y - (y/x))/(ln x)- (x/y))

At (3 , 6)

dy/dx = (ln 6 - (6/3))/(ln 3- (3/6))

=> (ln 6 - 2)/(ln 3 - 1/2)

=> 2*(ln 6 - 2)/(2*ln 3 - 1)

The required slope of the tangent is 2*(ln 6 - 2)/(2*ln 3 - 1)

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