Question

Find y'(
`\sqrt[3]{3}`) where y = y(x) is an implicit function of x given by 2
`x^(3)`+
`(x^(3)-3)`·sin(2πy)−3y = 0.

Answer 1

Given

2x³ + (x³ - 3) sin(2πy) - 3y = 0

we first notice that when x = ³√3, we get

2 (³√3)³ + ((³√3)³ - 3) sin(2πy) - 3y = 0

2•3 + (3 - 3) sin(2πy) - 3y = 0

6 - 3y = 0

3y = 6

y = 2

Differentiating both sides with respect to x gives

6x² + 3x³ sin(2πy) + 2π (x³ - 3) cos(2πy) y' - 3y' = 0

Then when x = ³√3, we find

6(³√3)² + 3(³√3)³ sin(2π•2) + 2π ((³√3)³ - 3) cos(2π•2) y' - 3y' = 0

6•³√9 + 3•3 sin(4π) + 2π (3- 3) cos(4π) y' - 3y' = 0

6•³√9 + 0 + 0 - 3y' = 0

3y' = 6•³√9

y' = 2•³√9

(that is, 2 times the cube root of 9)