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Linear approximation help?

+1 vote
let y = tan(4x+5)

find differential dy when x = 2 and dx = .4
find differential dy when x = 2 and dx = .8
asked Feb 21, 2013 in CALCULUS by andrew Scholar

1 Answer

+1 vote

y = tan(4x + 5)

Apply derivative each side.

dy = d[tan(4x + 5)]

Derivative of Trigonometric Functions: (d/dx)[tan(x)] = sec2(x)

dy = sec2(4x + 5)(4)dx                                                        inner derivative of (4x + 5) is 4

dy = 4sec2(4x + 5)dx

Reciprocal identities: sec(x) = 1/cos(x)

dy = [4/cos2(4x+5)]dx

Substituting x = 2 and dx = 0.4 in the equation.

dy = [4/cos2(4(2)+5)](0.4) = [4/cos2(8+5)](0.4) = 4(0.4)/[cos(13)]2

Trigonometric table in cos13 = 0.974

dy = (1.6)/[0.974]2 = 1.6/0.948676 = 1.68656.

Therefore x = 2 and dx = .4 then dy = 1.68656

Substituting x = 2 and dx = 0.8 in the equation.

dy = [4/cos2(4(2)+5)](0.8) = [4/cos2(8+5)](0.8) = 4(0.8)/[cos(13)]2

dy = (3.2)/[0.974]2 = 3.2/0.948676 = 3.3731221.

Therefore x = 2 and dx = .8 then dy = 3.3731221.

answered Feb 21, 2013 by britally Apprentice

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