Calculus: Derivatives Practice (Power, Product, Quotient and Chain Rules)

Power rule term by term: , , and the derivative of the constant 7 is 0. Leaving the +7 is a common slip.

Write . Power rule: .

Product rule: with and : . Multiplying the two derivatives () is the classic wrong shortcut.

Quotient rule: . The order matters (low d-high minus high d-low); reversing it gives the negative answer.

Chain rule: outer derivative times inner derivative 3 gives . Forgetting the inner derivative leaves .

Outer function sin becomes cos of the same inside, . Then multiply by the derivative of the inside, . The inside stays ; only the factor out front comes from differentiating it.

, so . The power rule does not apply to exponentials, so is wrong. is the antiderivative.

, with and . forgets the chain rule.

The point is (3, 9) and the slope is . Point-slope: , so . Check that the line passes through (3, 9): .

gives . f' changes from + to − at x = −1, so that is a local max with f(−1) = 2. At x = 1 it changes from − to +, a local min. is a zero of f, not a critical point.

, so the graph is concave up there, which makes it a local minimum, with f(1) = −2. The test is inconclusive only when ; the inflection point here is at x = 0.

Differentiate implicitly: , so . Sanity check: in the first quadrant the upper semicircle slopes downward, so the slope must be negative.

, so . is the volume itself, not its rate of change.

At rest means velocity is 0: , so t = 1 and t = 3. t = 0 and t = 3 are when the position is 0, which is not the same thing. t = 2 is when acceleration is 0.

Using the quotient rule on : . is the derivative of , and is the derivative of .

False. is continuous at 0 but has a sharp corner there, so it has no derivative. The converse is true: differentiable at a point implies continuous there.

Rewrite as and use , which gives 3. L'Hôpital's rule agrees: .