AP Calculus AB Product Rule, Quotient Rule and Trig Derivatives Quiz: 10 Questions with Explanations

A \(e^{x}\)B \(xe^{x}\)C \(e^{x}(x+1)\)D \(e^{x}+x\)

Product rule: 1·e^x + x·e^x = e^x(x + 1). Differentiating each factor separately gives only e^x.

A \(2x\cos x\)B \(2x\sin x-x^{2}\cos x\)C \(x^{2}\cos x\)D \(2x\sin x+x^{2}\cos x\)

Product rule: (2x)(sin x) + (x^2)(cos x). The minus sign belongs to the quotient rule, not the product rule.

A \(\frac{u^{\prime}v-uv^{\prime}}{v^{2}}\)B \(\frac{u^{\prime}v+uv^{\prime}}{v^{2}}\)C \(\frac{uv^{\prime}-u^{\prime}v}{v^{2}}\)D \(\frac{u^{\prime}}{v^{\prime}}\)

The numerator is low d-high minus high d-low, so u′v – uv′, over v squared.

A \(0\)B \(1\)C \(-1\)D \(\frac{1}{2}\)

f′(x) = (1 – x^2)/(x^2 + 1)^2, so f′(0) = 1/1 = 1.

A \(0\)B \(-1\)C \(1\)D \(\text{undefined}\)

The derivative is sec^2 x, and sec^2 0 = 1. The graph of tan x rises through the origin with slope 1.

A \(\sec^{2}x\)B \(\sec x\tan x\)C \(-\sec x\tan x\)D \(\csc x\cot x\)

The derivative of sec x is sec x tan x. Sec^2 x is the derivative of tan x, and the negative forms belong to the co-functions.

A \(\csc^{2}x\)B \(-\sec^{2}x\)C \(\tan x\)D \(-\csc^{2}x\)

The derivative of cot x is -csc^2 x. Co-functions have negative derivatives.

A \(0\)B \(-\pi\)C \(\pi\)D \(1\)

f′(x) = sin x + x cos x, so f′(pi) = 0 + pi(-1) = -pi. The value 0 is f(pi).

A \(18\)B \(-5\)C \(10\)D \(-2\)

Product rule: f′g + fg′ = 5(4) + 2(-1) = 18. Multiplying the derivatives gives -5, which is wrong.

A \(-1\)B \(1\)C \(-\frac{1}{3}\)D \(-3\)

(f′g – fg′)/g^2 = (1·3 – 6·2)/9 = -9/9 = -1. Swapping the order of the numerator gives +1.