How do you find the equation of the tangent line to each of the following curves at the given point: #(x^2)y=x+2# at (2,1)?

1 Answer
Bill K.
Jun 8, 2015

First note that the point #(2,1)# is on the graph of the equation #x^2y=x+2# since #2^2*1=2+2#.

The equation #x^2y=x+2# implies that #y=x^{-1}+2x^{-2}# when #x!=0#. The derivative is #dy/dx=-x^{-2}-4x^{-3}# so that the slope of the tangent line at #(x,y)=(2,1)# is #-2^{-2}-4*2^{-3}=-1/4-1/2=-3/4#.

The equation of the tangent line is then #y=f(2)+f'(2)(x-2)=1-3/4*(x-2)#, or #y=-3/4x+5/2#.

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