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

1 Answer
Jun 2, 2018

10

Explanation:

Taking the logarithm of both sides gives us

#ln y -ln x +x/y-x^2 = ln C#

Differentiating with respect to #x# yields

#1/y dy/dx -1/x +1/y-x/y^2dy/dx-2x = 0 implies#

#(1/y-x/y^2)dy/dx =2x+1/x-1/y implies #

#x(y-x)dy/dx = (2x^2+1)y^2-xy#

At the point #(1,2)# we have

#1 times (2-1)times dy/dx = (2 times1^2+1)times 2^2-1times 2 implies#
# dy/dx = 10#

Hence the slope of the tangent at #(1,2)# is 10

Note

I have assumed that tangent at #(1,2)# means tangent to the curve at #(1,2)#. Of course, this means that the constant #C# can not be arbitrary - but rather has to be #2/sqrt e#