Master derivatives of products and quotients with 10 comprehensive practice problems and step-by-step solutions.
When to use: When differentiating a product of two functions (both depend on x).
Example: d/dx [x²·sin(x)] = 2x·sin(x) + x²·cos(x)
When to use: When differentiating a quotient of two functions.
Example: d/dx [sin(x)/x] = [x·cos(x) - sin(x)]/x²
Find f'(x) using the product rule.
1 Identify f(x) and g(x):
f(x) = x³, g(x) = eˣ
2 Find derivatives:
f'(x) = 3x², g'(x) = eˣ
3 Apply product rule:
f'(x)·g(x) + f(x)·g'(x) = (3x²)·eˣ + (x³)·eˣ
4 Factor:
= eˣ(3x² + x³) = x²eˣ(3 + x)
5 Final answer:
f'(x) = x²eˣ(3 + x)
Find h'(x) using the quotient rule.
1 Identify numerator and denominator:
f(x) = x² + 1, g(x) = x - 1
2 Find derivatives:
f'(x) = 2x, g'(x) = 1
3 Apply quotient rule:
[f'(x)·g(x) - f(x)·g'(x)] / [g(x)]²
= [(2x)(x-1) - (x²+1)(1)] / (x-1)²
4 Expand numerator:
= [2x² - 2x - x² - 1] / (x-1)²
= [x² - 2x - 1] / (x-1)²
5 Final answer:
h'(x) = (x² - 2x - 1) / (x-1)²
Find dy/dx. (Hint: Use product rule twice or trigonometric identity first)
Method 1: Using trig identity (simpler)
1 Use identity: sin(x)cos(x) = (1/2)sin(2x)
y = x·(1/2)sin(2x) = (1/2)x·sin(2x)
2 Apply product rule:
dy/dx = (1/2)[1·sin(2x) + x·2cos(2x)]
= (1/2)sin(2x) + x·cos(2x)
Method 2: Double product rule
1 Let u = x·sin(x), v = cos(x)
2 First find u':
u' = 1·sin(x) + x·cos(x) = sin(x) + x·cos(x)
3 Apply product rule to y = u·v:
dy/dx = u'·v + u·v'
= [sin(x) + x·cos(x)]·cos(x) + [x·sin(x)]·[-sin(x)]
4 Simplify:
= sin(x)cos(x) + x·cos²(x) - x·sin²(x)
= (1/2)sin(2x) + x[cos²(x) - sin²(x)]
= (1/2)sin(2x) + x·cos(2x) ✓ (Same as Method 1)
Find dy/dx. Note: Denominator requires chain rule.
1 Rewrite denominator:
√(x² + 1) = (x² + 1)^(1/2)
2 Identify f(x) and g(x):
f(x) = eˣ, g(x) = (x² + 1)^(1/2)
3 Find derivatives:
f'(x) = eˣ
g'(x) = (1/2)(x² + 1)^(-1/2)·(2x) = x/√(x² + 1)
4 Apply quotient rule:
dy/dx = [eˣ·√(x²+1) - eˣ·(x/√(x²+1))] / (x²+1)
5 Multiply numerator and denominator by √(x²+1):
= [eˣ(x²+1) - eˣ·x] / [(x²+1)^(3/2)]
6 Factor eˣ:
= eˣ[x² + 1 - x] / (x²+1)^(3/2)
= eˣ(x² - x + 1) / (x²+1)^(3/2)
Find dy/dx. Derive the triple product rule formula.
Deriving triple product rule:
For y = f(x)·g(x)·h(x), let u = f(x)·g(x)
y = u·h(x)
dy/dx = u'·h(x) + u·h'(x)
But u' = f'(x)·g(x) + f(x)·g'(x)
∴ dy/dx = [f'(x)·g(x) + f(x)·g'(x)]·h(x) + f(x)·g(x)·h'(x)
= f'(x)·g(x)·h(x) + f(x)·g'(x)·h(x) + f(x)·g(x)·h'(x)
Applying to our problem:
1 Identify:
f(x) = x, g(x) = ln(x), h(x) = eˣ
2 Derivatives:
f'(x) = 1, g'(x) = 1/x, h'(x) = eˣ
3 Apply triple product rule:
dy/dx = (1)·ln(x)·eˣ + x·(1/x)·eˣ + x·ln(x)·eˣ
= eˣ[ln(x) + 1 + x·ln(x)]
= eˣ[ln(x)(1 + x) + 1]
| Rule | Formula | When to Use | Common Mistake |
|---|---|---|---|
| Product Rule | (fg)' = f'g + fg' | Two functions multiplied | Forgetting to add both terms |
| Quotient Rule | (f/g)' = (f'g - fg')/g² | One function divided by another | Forgetting to square denominator |
| Triple Product | (fgh)' = f'gh + fg'h + fgh' | Three functions multiplied | Missing one of the three terms |