3. Definition of the Derivative (part 2) For problems (a)-(d) below, \( a, b \) and \( c \) are constant. (a) Use the definition of the derivative to show that the derivative of the parabola \( f(x)= \) \( a x^{2}-b x+c \) is \( f^{\prime}(x)=2 a x-b \). (b) Use the definition of the derivative to show that the derivative of a cubic function \( f(x)=a x^{3} \) is \( f^{\prime}(x)=3 a x^{2} \). (c) Use the definition of the derivative to show that the derivative of the function \( f(x)=\sqrt{a x} \) is \( f^{\prime}(x)=\frac{a}{2 \sqrt{a x}} \). (d) Use the definition of the derivative to show that the derivative of the function \( f(x)=\frac{a}{b x} \) is \( f^{\prime}(x)=-\frac{a}{b x^{2}} \). (e) Verify your solutions to (a)-(d) using the Derivative Rules (Power Rule, Constant Multiple Rule, etc.).

We are going to use the definition   f′(x) = limₕ→0 [f(x + h) − f(x)]/h to differentiate each function. Finally, we’ll verify our answers using the standard derivative rules. ───────────────────────────── (a) f(x) = ax² – bx + c Step 1. Write f(x + h):   f(x + h) = a(x + h)² – b(x + h) + c Step 2. Expand:   a(x + h)² = a(x² + 2xh + h²) = ax² + 2axh + ah²   – b(x + h) = –bx – bh Thus, f(x + h) = ax² + 2axh + ah² – bx – bh + c Step 3. Compute the difference:   f(x + h) – f(x) = [ax² + 2axh + ah² – bx – bh + c] – [ax² – bx + c]     = 2axh + ah² – bh Step 4. Form the difference quotient:   [f(x + h) − f(x)]/h = (2axh + ah² – bh)/h     = 2ax + ah – b   (for h ≠ 0) Step 5. Take the limit as h → 0:   f′(x) = limₕ→0 [2ax + ah – b] = 2ax – b ───────────────────────────── (b) f(x) = ax³ Step 1. Write f(x + h):   f(x + h) = a(x + h)³ Step 2. Expand (x + h)³:   (x + h)³ = x³ + 3x²h + 3xh² + h³ Thus, f(x + h) = a(x³ + 3x²h + 3xh² + h³) = ax³ + 3ax²h + 3axh² + ah³ Step 3. Difference:   f(x + h) – f(x) = [ax³ + 3ax²h + 3axh² + ah³] – ax³ = 3ax²h + 3axh² + ah³ Step 4. Difference quotient:   [f(x + h) − f(x)]/h = (3ax²h + 3axh² + ah³)/h     = 3ax² + 3axh + ah² Step 5. Let h → 0:   f′(x) = limₕ→0 [3ax² + 3axh + ah²] = 3ax² ───────────────────────────── (c) f(x) = √(ax)    or   (ax)^(1/2) Step 1. Write f(x + h):   f(x + h) = √(a(x + h)) = √(ax + ah) Step 2. Form the difference quotient:   [f(x + h) − f(x)]/h = [√(ax + ah) − √(ax)] / h Step 3. Multiply by the conjugate over itself to simplify:   = { [√(ax + ah) − √(ax)] · [√(ax + ah) + √(ax)] } / { h [√(ax + ah) + √(ax)] }   The numerator becomes:    (ax + ah) – ax = ah Thus, the expression simplifies to:   = [ah] / [h (√(ax + ah) + √(ax))]   Cancel h (assuming h ≠ 0):   = a / [√(ax + ah) + √(ax)] Step 4. Take the limit as h → 0:   As h → 0, √(ax + ah) → √(ax).   Thus, limit becomes:   f′(x) = a / [2√(ax)]   or   f′(x) = a/(2√(ax)) ───────────────────────────── (d) f(x) = a/(bx)    which can also be written as  (a/b)(1/x) Step 1. Write f(x + h):   f(x + h) = a/(b(x + h)) Step 2. Form the difference quotient:   [f(x + h) − f(x)]/h = [a/(b(x + h)) − a/(bx)] / h Step 3. Combine the two fractions:   Find a common denominator for the terms in brackets:   Common denominator = b x(x + h)   Thus, a/(b(x + h)) − a/(bx) = [a x – a(x + h)] / [b x(x + h)]     = [a x – a x – a h] / [b x(x + h)] = (–a h) / [b x(x + h)] Step 4. Divide by h:   [f(x + h) − f(x)]/h = (–a h) / [b x(x + h) · h] = –a / [b x(x + h)] Step 5. Take the limit as h → 0:   As h → 0, (x + h) → x, so   f′(x) = –a / [b x²] ───────────────────────────── (e) Verification Using Derivative Rules Let’s quickly check each result with known rules: 1. For (a) f(x) = ax² – bx + c, using the power rule and constant multiple rule:   d/dx(ax²) = 2ax, d/dx(–bx) = –b, d/dx(c) = 0   Thus, f′(x) = 2ax – b. 2. For (b) f(x) = ax³:   d/dx(ax³) = a · 3x² = 3ax². 3. For (c) f(x) = (ax)^(1/2):   Using the chain rule: d/dx [ (ax)^(1/2) ] = (1/2)(ax)^(–1/2) · a = a/(2√(ax)). 4. For (d) f(x) = a/(bx) = (a/b)x^(–1):   d/dx[(a/b)x^(–1)] = (a/b)(–1)x^(–2) = –a/(b x²). ───────────────────────────── Summary of Answers: (a) f′(x) = 2ax – b (b) f′(x) = 3ax² (c) f′(x) = a/(2√(ax)) (d) f′(x) = –a/(b x²) All our results match the verification using the derivative rules.