5x2y - 3xy2= 14. Show that the point (2,1) is on this graph. Find an expression for dy/dx and find the value of the slope at (2,1)

Question

5x2y - 3xy2= 14. Show that the point (2,1) is on this graph. Find an expression for dy/dx and find the value of the slope at (2,1)

Answer 1

5x²y - 3xy² = 14

 

(2,1):

 

5(2)²(1) - 3(2)(1)² = 14

5(4)(1) - 3(2)(1) = 14

20 - 6 = 14

14 = 14

 

d/dx (5x²y - 3xy² = 14)

 

5x²(dy/dx) + 10xy - [3x(2y dy/dx) + 3y²] = 0

 

5x²(dy/dx) + 10xy - 6xy(dy/dx) - 3y² = 0

 

5x²(dy/dx) - 6xy(dy/dx) = 3y² - 10xy

 

(5x² - 6xy)(dy/dx) = 3y² - 10xy

 

dy/dx = (3y² - 10xy)/(5x² - 6xy)

 

dy/dx (2,1) = [3(1)² - 10(2)(1)]/[5(2)² - 6(2)(1)]

 

dy/dx (2,1) = [3 - 20]/[5(4)-12] = -17/(20-12) = -17/8