Differentiate y = (2x + 3)^3 and evaluate dy/dx at x = -1.
Answer and explanation
Correct answer: B. 6
- Step 1: Apply the chain rule: dy/dx = 3 x 2 x (2x + 3)^2.
- Step 2: At x = -1 the inner expression equals 1.
- Step 3: dy/dx = 3 x 2 x 1^2 = 6.
Answer
6.
Why not the others: '3' omits the inner derivative 2 from the chain rule; '1' evaluates y instead of dy/dx; '2' forgets to bring down the exponent 3.
Practise more Calculus and Analytical Geometry.
Free questions with an explanation for every one, and full timed Qatar Senior Secondary Certificate (Thanaweya) mock tests.
More Calculus and Analytical Geometry questions
- Evaluate the limit: lim_(x -> 0) [sin(5x) / (2x)].
- Find the derivative dy/dx of the implicit equation x^3 + y^3 = 6xy at the point (3, 3) on the Folium of Descartes.
- What is the equation of the normal line to the curve y = x^2 - 4x + 5 at the point (1, 2)?
- Evaluate the definite integral: integral from 0 to 1 of [x / (x^2 + 1)] dx.
- Find the absolute maximum value of the function f(x) = 2x^3 - 3x^2 - 12x + 1 on the closed interval [-2, 3].
- What is the total area of the finite region enclosed between the parabola y = 4 - x^2 and the line y = 0?