AP Calculus AB The Chain Rule Quiz: 10 Questions with Explanations

A \(4(2x+1)^{3}\)B \(8(2x+1)^{3}\)C \(4(2x)^{3}\)D \(8(2x+1)^{4}\)

Outer derivative 4(2x + 1)^3 times the inner derivative 2 gives 8(2x + 1)^3. Forgetting the 2 gives 4(2x + 1)^3.

A \(3\cos 3x\)B \(\cos 3x\)C \(-3\cos 3x\)D \(3\sin 3x\)

The derivative of sine is cosine, and the inner function 3x contributes a factor 3.

A \(e^{x^{2}}\)B \(2xe^{2x}\)C \(x^{2}e^{x^{2}-1}\)D \(2xe^{x^{2}}\)

The exponential stays and is multiplied by the inner derivative 2x. Using the power rule on the exponent is a common mistake.

A \(\frac{1}{x^{2}+4}\)B \(\frac{2x}{x^{2}}\)C \(\frac{2x}{x^{2}+4}\)D \(2x\ln(x^{2}+4)\)

The derivative of ln u is u′/u, so the result is 2x/(x^2 + 4).

A \(0\)B \(1\)C \(2\)D \(-2\)

The derivative is 2cos 2x, which is 2 at x = 0. Without the chain factor, the slope would be 1.

A \(\frac{x}{\sqrt{x^{2}+9}}\)B \(\frac{1}{2\sqrt{x^{2}+9}}\)C \(\frac{2x}{\sqrt{x^{2}+9}}\)D \(\frac{x}{2\sqrt{x^{2}+9}}\)

The derivative is (1/2)(x^2 + 9)^(-1/2) times 2x, and the 2s cancel to leave x/sqrt(x^2 + 9).

A \(3\)B \(12\)C \(4\)D \(7\)

f′(g(2)) · g′(2) = f′(5)·3 = 12. Adding the derivatives gives 7, and using f′(2) instead of f′(5) is another trap.

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

Outer: u^2 gives 2 cos x, inner cos x gives -sin x. The product is -2 cos x sin x.

A \(4\)B \(8\)C \(2\)D \(16\)

y′ = 2(x^2 + 1)(2x) = 4x(x^2 + 1), which is 8 at x = 1.

A \(e^{\cos x}\)B \(\sin x\,e^{\cos x}\)C \(-\sin x\,e^{x}\)D \(-\sin x\,e^{\cos x}\)

Keep e^(cos x) and multiply by the derivative of cos x, which is -sin x.