y = tan (½ tan-¹ x)

tan-¹ y = ½ tan-¹ x

Differentiate both sides with respect to x,

(1/(1+y²)) dy/dx = ½ (1/(1+x²))

dy/dx = ½(1+y²)(1/(1+x²))

(1+x²) dy/dx = ½(1+y²)

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(1+x²) dy/dx = ½(1+y²) ①

Differentiate with respect to x again,

(1+x²) d²y/dx² + 2x dy/dx = ½(2y dy/dx)

(1+x²) d²y/dx² + 2x dy/dx = y dy/dx ②


Differentiate with respect to x again,

(1+x²) d³y/dx³ + 2x d²y/dx² + 2 dy/dx + 2x d²y/dx² = dy/dx (dy/dx) + y d²y/dx² ③


For Maclaurin's series at x = 0,

y = tan(½ tan-¹ 0)
y = tan (½ x 0)
y = tan 0 = 0

Looking at ①, when x = 0, y = 0,

(1+0²) dy/dx = ½(1+0²)

dy/dx = ½


Looking at ②, when x = 0, y = 0, dy/dx = ½,

(1+0²) d²y/dx² + 2(0)(½) = (0)(½)

d²y/dx² = 0


Looking at ③, when x = 0, y = 0, dy/dx = ½, d²y/dx² = 0,


(1+0²) d³y/dx³ + 2(0)(0) + 2(½) + 2(0)(0) = (½)(½) + 0(0)

d³y/dx³ + 1 = ¼
d³y/dx³ = -¾



For Maclaurin series,

f(x) = f(0) + x f'(0) + x²/2! f''(0) + x³/3! f'''(0) + ...

So y = tan(½ tan-¹ x)

= 0 + x (½) + x²/2! (0) + x³/3! (-¾) + ...
= ½x + x³/6 (-¾) + ...
= ½x - ⅛x³ + ...
≈ ½x - ⅛x³

(Shown)