Consider the relation defined by the equation tan(y)= x + y for x in the open interval -2 pi < x < 2 pi?
Consider the relation defined by the equation ( \tan(y) = x + y ) for ( x ) in the open interval ( -2\pi < x < 2\pi ).
a. Find ( \frac{dy}{dx} ) in terms of ( y ).
b. Determine the coordinates of each point where the tangent line to the graph is vertical.
c. Calculate ( \frac{d^2y}{dx^2} ) in terms of ( y ).
1 Answers
a.
Implicit differentiation:
sec²y (dy/dx) = 1 + dy/dx
(sec²y - 1)(dy/dx) = 1
tan²y (dy/dx) = 1
dy/dx = 1/tan²y
dy/dx = cot²y
b.
dy/dx is undefined at y = nπ, where n is an integer.
tan(nπ) = x + nπ
0 = x + nπ
x = -nπ
Solutions satisfying -2π < x < 2π:
(-π, π), (0, 0), (π, -π)
c.
dy/dx = cot²y
d²y/dx² = 2cot(y)(csc²y)(dy/dx)
d²y/dx² = 2cot(y)(csc²y)(cot²y)
d²y/dx² = 2cot³y csc²y
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