How do you find #(dy)/(dx)# given #(2x-3)^2+(4y-5)^2=10#?

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Answer 1 · 2016-11-06T13:56:37

The derivative is #dy/dx = (3 - 2x)/(8y - 10)#.

Start by expanding the parentheses.

#4x^2 - 12x + 9 + 16y^2 - 40y + 25 = 10#

#d/dx(4x^2 - 12x + 9 + 16y^2 - 40y + 25) = d/dx(10)#

#d/dx(4x^2) + d/dx(-12x) + d/dx(9) + d/dx(16y^2) + d/dx(-40y) + d/dx(25) = d/dx(10)#

#8x - 12 + 0 + 32y(dy/dx) - 40(dy/dx) + 0 = 0#

#32y(dy/dx) - 40(dy/dx) = 12 - 8x#

#dy/dx(32y - 40) = 12 - 8x#

#dy/dx = (12 - 8x)/(32y - 40)#

#dy/dx = (4(3 - 2x))/(4(8y - 10))#

#dy/dx= (3- 2x)/(8y - 10)#

Hopefully this helps!