Solutions — Differentiation from First Principles

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1. Q1 — Constants and linear functions

   (a) f(x)=7. f(x+h)=7. Numerator: 7−7=0. Quotient: 0/h=0. f′(x)=0.   (b) f(x)=4x. f(x+h)=4x+4h. Numerator: 4h. Quotient: 4. f′(x)=4.   (c) f(x+h)=2x+2h+9. Numerator: 2h. Quotient: 2. f′(x)=2.   (d) f(x+h)=−3x−3h+1. Numerator: −3h. Quotient: −3. f′(x)=−3.

2. Q2 — Quadratics (x² form)

   (a) f(x+h)=(x+h)²=x²+2xh+h². Num: 2xh+h²=h(2x+h). Quotient: 2x+h. f′(x)=2x.   (b) f(x+h)=3(x+h)²=3x²+6xh+3h². Num: 6xh+3h²=h(6x+3h). Quotient: 6x+3h. f′(x)=6x.   (c) f(x+h)=−(x+h)²=−x²−2xh−h². Num: −2xh−h²=h(−2x−h). Quotient: −2x−h. f′(x)=−2x.   (d) f(x+h)=(x+h)²+5=x²+2xh+h²+5. Num: 2xh+h²=h(2x+h). Quotient: 2x+h. f′(x)=2x.

3. Q3 — Quadratics (mixed)

   (a) f(x+h)=x²+2xh+h²+3x+3h. Num: 2xh+h²+3h=h(2x+h+3). Quotient: 2x+h+3. f′(x)=2x+3.   (b) f(x+h)=2x²+4xh+2h²−4x−4h+1. Num: 4xh+2h²−4h=h(4x+2h−4). Quotient: 4x+2h−4. f′(x)=4x−4.   (c) f(x+h)=−x²−2xh−h²+6x+6h−2. Num: −2xh−h²+6h=h(−2x−h+6). Quotient: −2x−h+6. f′(x)=−2x+6.

4. Q4 — Evaluating the derivative

   (a) f′(x)=2x. f′(3)=6.   (b) f′(x)=2. f′(5)=2.   (c) f′(x)=2x+4. f′(−1)=−2+4=2.   (d) f′(x)=−4x. f′(2)=−8.

5. Q5 — First principles for cubics

   (a) f(x+h)=(x+h)³=x³+3x²h+3xh²+h³. Num: 3x²h+3xh²+h³=h(3x²+3xh+h²). Quotient: 3x²+3xh+h². lim: f′(x)=3x² ✓   (b) At (2,8): f′(2)=3(4)=12. At (−1,−1): f′(−1)=3(1)=3.   (c) 3x²=12 → x²=4 → x=±2. Points: (2,8) and (−2,−8).

6. Q6 — Interpreting the derivative

   (a) f(x)=x²−6x+9. f′(x)=2x−6. f′(3)=0: the tangent at (3,0) is horizontal; (3,0) is the vertex.   (b) g′(x)=4x+1. Set 4x+1=5 → 4x=4 → x=1.   (c) If h′(x)=4x−2 came from h(x)=ax²+bx+c by first principles, then 2a=4⇒a=2 and b=−2. One answer: h(x)=2x²−2x (c can be any constant).

7. Q7 — Tangent equation

   (a) f′(x)=2x+2. f′(3)=8. Gradient = 8.   (b) y−15=8(x−3) → y=8x−24+15 → y=8x−9.   (c) f′(x)=0 → 2x+2=0 → x=−1. The tangent is horizontal at x=−1, at the point (−1,−1).

8. Q8 — General coefficients

   (a) f(x+h)=a(x+h)+b=ax+ah+b. Num: ah. Quotient: a. lim: f′(x)=a. Independent of x — confirms linear functions have constant gradient.   (b) f(x+h)=a(x+h)²=ax²+2axh+ah². Num: 2axh+ah²=h(2ax+ah). Quotient: 2ax+ah. lim: f′(x)=2ax.   (c) Using (a) and (b): d/dx(5x²)=10x, d/dx(−3x)=−3, d/dx(8)=0. So f′(x)=10x−3.

9. Q9 — Non-standard function

   (a) f(x+h)−f(x)=(x+h)²+(x+h)³−x²−x³=(2xh+h²)+(3x²h+3xh²+h³)=h(2x+h+3x²+3xh+h²).   (b) Quotient: 2x+h+3x²+3xh+h². lim: f′(x)=2x+3x².   (c) 3x²+2x=5 → 3x²+2x−5=0 → (3x+5)(x−1)=0 → x=1 or x=−5/3.

10. Q10 — Parabola geometry

    (a) f(x+h)−f(x)=(2xh+h²)+(h)(−4)=(2xh+h²−4h). Wait, full: f(x+h)=x²+2xh+h²−4x−4h+3. Num: 2xh+h²−4h=h(2x+h−4). Quotient: 2x+h−4. y′=2x−4.   (b) 2x−4=−2 → x=1. y=1−4+3=0. Point: (1, 0).   (c) y−0=−2(x−1) → y=−2x+2.   (d) Vertex where y′=0: 2x−4=0 → x=2. Check: y=x²−4x+3=(x−2)²−1 ✓. Vertex at (2,−1).