Solutions — Chain Rule for Trig, Exponential and Log Functions

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1. Show Solution
   y = cos(2x + 1)
   Outer function: cos( ); inner function: 2x + 1; inner derivative: 2.
   dy/dx = −2 sin(2x + 1)
2. Show Solution
   y = sin(x²)
   Outer function: sin( ); inner function: x²; inner derivative: 2x.
   dy/dx = 2x cos(x²)
3. Show Solution
   y = e^{sin x}
   Outer function: e^{( )}; inner function: sin x; inner derivative: cos x.
   dy/dx = cos x · e^{sin x}
4. Show Solution
   y = ln(tan x)
   Outer function: ln( ); inner function: tan x; inner derivative: sec²x.
   dy/dx = sec²x / tan x
   Since sec²x = 1/cos²x and tan x = sin x/cos x:
   dy/dx = (1/cos²x) ÷ (sin x/cos x) = 1/(cos x · sin x) = 1/(sin x cos x)
5. Show Solution
   y = sin³x = (sin x)³
   Outer function: ( )³; inner function: sin x; inner derivative: cos x.
   dy/dx = 3(sin x)² · cos x = 3 sin²x cos x
6. Show Solution
   y = cos⁴(3x) = (cos(3x))⁴
   This requires two applications of the chain rule (or the power-chain rule applied carefully):
   Step 1: Outer power rule → 4(cos(3x))³
   Step 2: Derivative of cos(3x) = −3 sin(3x)
   dy/dx = 4(cos(3x))³ × (−3 sin(3x)) = −12 cos³(3x) sin(3x)
7. Show Solution
   y = e^2x+1
   Outer function: e^{( )}; inner function: 2x + 1; inner derivative: 2.
   dy/dx = 2e^2x+1
   At x = 0: gradient = 2e²⁽⁰⁾⁺¹ = 2e¹ = 2e
8. Show Solution
   y = sin(2x)
   At x = π/4: y = sin(2 × π/4) = sin(π/2) = 1. Point: (π/4, 1).
   dy/dx = 2 cos(2x)
   At x = π/4: gradient = 2 cos(π/2) = 2 × 0 = 0
   Tangent: y − 1 = 0(x − π/4) ⇒ y = 1
   This is a horizontal tangent because x = π/4 is the maximum of sin(2x).
9. Show Solution
   y = ln(sin²x)
   Apply log law first: ln(sin²x) = 2 ln(sin x)
   Now differentiate: dy/dx = 2 × (cos x / sin x) = 2 cot x
   
   (Alternatively without log laws: inner function sin²x, inner derivative 2 sin x cos x;
   dy/dx = 2 sin x cos x / sin²x = 2 cos x / sin x = 2 cot x ✓)
10. Show Solution
    x(t) = e^−t sin(2t)
    Using the product rule with chain rule:
    • d/dt[e^−t] = −e^−t
    • d/dt[sin(2t)] = 2 cos(2t)
    v(t) = x′(t) = −e^−t sin(2t) + e^−t · 2 cos(2t) = e^−t[2 cos(2t) − sin(2t)]
    
    Setting v = 0: e^−t ≠ 0 for all t, so:
    2 cos(2t) − sin(2t) = 0
    tan(2t) = 2
    2t = arctan(2) + nπ (for integer n)
    First solution for t > 0: t = arctan(2)/2 ≈ 1.107/2 ≈ 0.554 s