What is the slope of the tangent line of $x^3y^2-(x+y)/(x-y)^2= C$ , where C is an arbitrary constant, at $(1,4)$ ?

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Answer 1 · 2018-01-20T08:55:23

Slope of tangent line is $-1303/203$

As we are seeking slope at $(1,4)$ , this point lies on the curve given by $x^3y^2-(x+y)/(x-y)^2=C$ .

Putting values from $(1,4)$ , we get $1^3*4^2-(1+4)/(1-4)^2=C$

or $C=16-5/9=139/9$ and function is $x^3y^2-(x+y)/(x-y)^2=139/9$

Now differentiating the imlicit function $x^3y^2-(x+y)/(x-y)^2=C$

$3x^2y^2+2x^3y(dy)/(dx)-(1+(dy)/(dx))/(x-y)^2-(2(x+y))/(x-y)^3(1-(dy)/(dx))=0$

or $3x^2y^2(x-y)^3+2x^3y(x-y)^3(dy)/(dx)-(x-y)(1+(dy)/(dx))-2(x+y)(1-(dy)/(dx))=0$

or $[2x^3y(x-y)^3-(x-y)+2(x+y)] (dy)/(dx)=[-3x^2y^2(x-y)^3+x-y+2(x+y)]$

or $(dy)/(dx)=[-3x^2y^2(x-y)^3+x-y+2(x+y)]/[2x^3y(x-y)^3-(x-y)+2(x+y)]$

= $[-3x^2y^2(x-y)^3+3x+y]/[2x^3y(x-y)^3+x+3y]$

and at $(1,4)$ , slope is $[-3*1^2*4^2(1-4)^3+3+4]/[2*1^3*4(1-4)^3+1+3*4]$

= $(48*27+7)/(-8*27+13)$

= $-1303/203$

The tangent at $(1,4)$ looks liike:

graph{(x^3y^2-(x+y)/(x-y)^2-139/9)(203y-812+1303x-1303)=0 [-10.87, 9.13, -4.08, 5.92]}

What is the slope of the tangent line of x^3y^2-(x+y)/(x-y)^2= C x^3y^2-(x+y)/(x-y)^2= C , where C is an arbitrary constant, at (1,4)(1,4)?