Composite, Implicit, and Inverse Functions - 15 Multiple Choice + 6 Free Response Questions
Recommended time: 45 minutes
Solution:
1) Let g(x) = x² + 2x (inside function)
2) Let f(u) = u⁵ (outside function)
3) g'(x) = 2x + 2
4) f'(u) = 5u⁴
5) Using chain rule: f'(x) = f'(g(x)) · g'(x)
6) f'(x) = 5(x² + 2x)⁴ · (2x + 2)
7) = 5(x² + 2x)⁴ · 2(x + 1)
8) = 10(x + 1)(x² + 2x)⁴
Answer: 10(x + 1)(x² + 2x)⁴
Solution:
1) Differentiate both sides with respect to x:
2) d/dx[x²y] + d/dx[y³] = d/dx[8]
3) Use product rule for x²y: 2xy + x² · dy/dx
4) Chain rule for y³: 3y² · dy/dx
5) So: 2xy + x² · dy/dx + 3y² · dy/dx = 0
6) Factor out dy/dx: dy/dx(x² + 3y²) = -2xy
7) Solve for dy/dx: dy/dx = -2xy/(x² + 3y²)
Answer: dy/dx = -2xy/(x² + 3y²)
Solution:
1) Let g(x) = 3x (inside function)
2) Let f(u) = arcsin(u) (outside function)
3) g'(x) = 3
4) f'(u) = 1/√(1 - u²)
5) Using chain rule: f'(x) = f'(g(x)) · g'(x)
6) f'(x) = 1/√(1 - (3x)²) · 3
7) = 3/√(1 - 9x²)
Answer: 3/√(1 - 9x²)
Solution:
1) Let u = e^(x² + 1) and v = ln(x² + 1)
2) Find u': u' = e^(x² + 1) · 2x = 2xe^(x² + 1)
3) Find v': v' = 2x/(x² + 1)
4) Using product rule: f'(x) = u'v + uv'
5) = 2xe^(x² + 1) · ln(x² + 1) + e^(x² + 1) · 2x/(x² + 1)
6) = e^(x² + 1)[2x ln(x² + 1) + 2x/(x² + 1)]
7) = 2xe^(x² + 1)[ln(x² + 1) + 1/(x² + 1)]
Answer: 2xe^(x² + 1)[ln(x² + 1) + 1/(x² + 1)]
Solution:
1) Differentiate both sides with respect to x:
2) 3x² + 3y² · dy/dx = 3y + 3x · dy/dx
3) Collect dy/dx terms: 3y² · dy/dx - 3x · dy/dx = 3y - 3x²
4) Factor: dy/dx(3y² - 3x) = 3y - 3x²
5) Factor out 3: dy/dx(3)(y² - x) = 3(y - x²)
6) Cancel 3: dy/dx(y² - x) = y - x²
7) Solve: dy/dx = (y - x²)/(y² - x)
Answer: dy/dx = (y - x²)/(y² - x)
Solution:
1) Let y = (x² + 1)^x
2) Take natural log: ln(y) = ln((x² + 1)^x) = x ln(x² + 1)
3) Differentiate implicitly: (1/y) · dy/dx = ln(x² + 1) + x · 2x/(x² + 1)
4) Simplify: (1/y) · dy/dx = ln(x² + 1) + 2x²/(x² + 1)
5) Solve for dy/dx: dy/dx = y[ln(x² + 1) + 2x²/(x² + 1)]
6) Substitute back: dy/dx = (x² + 1)^x[ln(x² + 1) + 2x²/(x² + 1)]
Answer: (x² + 1)^x[ln(x² + 1) + 2x²/(x² + 1)]