Question 1187871
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Please help me with my assignment.
Use Implicit differentiation to find the Derivative of y with respect to x.
1. Find the slope of a tangent line to th graph of x^2y^2-xy+x=1 at (1,1)?
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You should differentiate this identity.


To facilitate it, differentiate each addend separately.


    (x^2*y^2)' = 2x*y^2 + 2x^2*y*y' 

    (xy)' = y + xy'

    x' = 1

    1' = 0.


Then combine these derivatives in one formula


    2x*y^2 + 2x^2*y*y' - y - xy' + 1 = 0


Keep the terms with y' on the left side; move the rest of the terms to the right side


    2x^2*y*y' - xy' = 2x*y^2 + y - 1


In the left side, factor out the common factor y'


    (2x^2*y - x)*y' = 2x*y^2 + y - 1


Express y', dividing both sides by  (2x^2*y - x)


    y' = {{{(2x*y^2 + y - 1)/(2x^2*y - x)}}}.


Thus the formula is just ready.

To get the value,  substitute the values  x= 1, y= 1 into the formula.  You will get then


    y' = {{{(2*1*1^2 + 1 - 1)/(2*1^2*1 - 1)}}} = {{{(2 + 1 - 1)/(2-1)}}} = {{{2/1}}} = 2.    <U>ANSWER</U>


<U>ANSWER</U>.  y' = 2.
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Solved and explained in all details.