Linear approximation help?

Question

let y = tan(4x+5)

find differential dy when x = 2 and dx = .4
find differential dy when x = 2 and dx = .8

Answer 1

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.