So, 1 2 and 1 2В are the roots of the given quadratic equation. So, - 3 5 and 5В are the roots of the given quadratic equation. So, - 1 2 and - 13В are the roots of the given quadratic equation. So, 6 and 9 are the roots of the given quadratic equation. Since y'' < 0, the graph is concave down at the point (-1, 1). Now let x=-1, y=1, and y'=4/5 so that the second derivative is Now determine the concavity of the graph at (-1, 1). Since y'= 4/5, the slope of the graph is 4/5 and the graph is increasing at the point (-1, 1). Thus, the slope of the graph (the slope of the line tangent to the graph) at (-1, 1) is Now differentiate both sides of the original equation, gettingД ( x 2 y ) + D ( y 4 ) = D ( 4 ) + D ( 2 x ) , Thus, the slope of the line tangent to the graph at (1, 3) is Now differentiate both sides of the original equation, getting Thus, the slope of the line tangent to the graph at the point (-1, 1) isДЄnd the tangent line passes through the point (1, 3). (Remember to use the chain rule on D ( y 2 ). Now differentiate both sides of the equation, gettingДЈ ( x 2+ y 2) 2 D ( x 2+ y 2) = 8 x 2 D ( y 2 ) + D ( 8 x 2 ) y 2 , (Remember to use the chain rule on D ( y 4 ) and D ( y 7 ). Clear the fractions by multiplying both sides of the equation by x 3 y 3, gettingД ( y 4 ) + D ( x 4 ) = x 5 D ( y 7 ) + D ( x 5 ) y 7 , Clear the fraction by multiplying both sides of the equation by y + x 2, getting (Remember to use the chain rule on D ( y 3 ) and Differentiate both sides of the equation, getting (Remember to use the chain rule on D ( x- y) 2. Differentiate both sides of the equation, gettingД ( x- y) 2 = D ( x ) + D ( y ) - D ( 1 ) , )Д¬lick HERE to return to the list of problems. (Remember to use the chain rule on D ( y 3 ). SOLUTIONS TO IMPLICIT DIFFERENTIATION PROBLEMS Solutions to Implicit Differentiation Problems

4 x 1 3 x 0