How do you find the slope of the line tangent to the graph of #ln(xy)-x=0# at the point where x=-1?

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Answer 1 · 2016-11-21T00:25:51

The slope of the line tangent is #-2/e#.

Simplify the function and then differentiate.

#ln(xy) - x =0#

#ln(xy) = x#

#xy = e^x#

#y + x(dy/dx) = e^x#

#x(dy/dx) =e^x - y#

#dy/dx = (e^x- y)/x#

The slope of the tangent is given by evaluating the point #(x, y)# within the derivative.

We will need to find the y-coordinate of the point of contact.

#ln(-1y) - (-1) = 0#

#ln(-y) + 1 = 0#

#ln(-y) = -1#

#-y = e^-1#

#-y = 1/e#

#y = -1/e#

So,

#m_"tangent" = (e^-1 - (-1/e))/-1#

#m_"tangent" = (1/e + 1/e)/-1#

#m_"tangent" = (2/e)/-1#

#m_"tangent" = -2/e#

Hopefully this helps!