Calculus: Derivatives Worksheet

Question 1

Use the limit definition of the derivative to find f'(x) for f(x) = x^2.

Accepted Answer

Using f'(x) = lim h approaches 0 of [(x + h)^2 - x^2]/h, we get lim h approaches 0 of [2xh + h^2]/h = lim h approaches 0 of 2x + h = 2x. Therefore, f'(x) = 2x.

Question 2

Find the derivative of f(x) = 7x^5 - 3x^2 + 9x - 4.

Accepted Answer

The derivative is f'(x) = 35x^4 - 6x + 9. The constant term -4 has derivative 0.

Question 3

Find dy/dx if y = 4x^-3 + 6sqrt(x). Write your answer using exponents or radicals.

Accepted Answer

Rewrite the function as y = 4x^-3 + 6x^(1/2). The derivative is dy/dx = -12x^-4 + 3x^(-1/2), which can also be written as -12/x^4 + 3/sqrt(x).

Question 4

At x = 2, find the slope of the tangent line to the curve f(x) = x^3 - 4x + 1.

Accepted Answer

First find f'(x) = 3x^2 - 4. Then evaluate at x = 2: f'(2) = 3(2)^2 - 4 = 12 - 4 = 8. The slope of the tangent line is 8.

Question 5

Find the equation of the tangent line to f(x) = x^2 + 3x at x = 1.

Accepted Answer

The derivative is f'(x) = 2x + 3, so the slope at x = 1 is 5. The point on the graph is f(1) = 1 + 3 = 4, so the tangent line passes through (1, 4). Using point-slope form, y - 4 = 5(x - 1), which simplifies to y = 5x - 1.

Question 6

Find the derivative of g(x) = (2x^3 - x)(5x^2 + 1).

Accepted Answer

Use the product rule. Let u = 2x^3 - x and v = 5x^2 + 1. Then u' = 6x^2 - 1 and v' = 10x. So g'(x) = (6x^2 - 1)(5x^2 + 1) + (2x^3 - x)(10x). This simplifies to g'(x) = 50x^4 - 12x^2 - 1.

Question 7

Find the derivative of h(x) = (x^2 + 1)/(x - 3).

Accepted Answer

Use the quotient rule. h'(x) = [(x - 3)(2x) - (x^2 + 1)(1)]/(x - 3)^2. The numerator simplifies to 2x^2 - 6x - x^2 - 1 = x^2 - 6x - 1. Therefore, h'(x) = (x^2 - 6x - 1)/(x - 3)^2.

Question 8

Find the derivative of f(x) = (3x - 2)^5.

Accepted Answer

Use the chain rule. The outside derivative is 5(3x - 2)^4, and the inside derivative is 3. Therefore, f'(x) = 15(3x - 2)^4.

Question 9

Find the derivative of y = sin(x) + 4cos(x) - 2tan(x).

Accepted Answer

The derivative is dy/dx = cos(x) - 4sin(x) - 2sec^2(x). This uses the facts that the derivative of sin(x) is cos(x), the derivative of cos(x) is -sin(x), and the derivative of tan(x) is sec^2(x).

Question 10

Find the derivative of y = e^x ln(x), where x is greater than 0.

Accepted Answer

Use the product rule. Let u = e^x and v = ln(x). Then u' = e^x and v' = 1/x. So dy/dx = e^x ln(x) + e^x/x.

Question 11

A car's position is given by s(t) = t^3 - 6t^2 + 9t, where s is in meters and t is in seconds. Find the car's velocity function.

Accepted Answer

Velocity is the derivative of position. Since s(t) = t^3 - 6t^2 + 9t, the velocity function is v(t) = s'(t) = 3t^2 - 12t + 9 meters per second.

Question 12

For the car in the previous problem, find the acceleration function.

Accepted Answer

Acceleration is the derivative of velocity. Since v(t) = 3t^2 - 12t + 9, the acceleration function is a(t) = v'(t) = 6t - 12 meters per second squared.

Question 13

The graph of f(x) is shown. Estimate f'(2) from the tangent line. The tangent line appears to pass through the points (1, 3) and (3, 7).

Accepted Answer

The derivative f'(2) is the slope of the tangent line at x = 2. Using the points (1, 3) and (3, 7), the slope is (7 - 3)/(3 - 1) = 4/2 = 2. Therefore, f'(2) is approximately 2.

Question 14

A function has derivative f'(x) = 3x^2 - 12x + 9. Find the x-values where the graph of f has horizontal tangent lines.

Accepted Answer

Horizontal tangent lines occur where f'(x) = 0. Solve 3x^2 - 12x + 9 = 0 by factoring: 3(x^2 - 4x + 3) = 0, so 3(x - 1)(x - 3) = 0. Therefore, the graph has horizontal tangent lines at x = 1 and x = 3.

Question 15

Use implicit differentiation to find dy/dx for x^2 + y^2 = 25.

Accepted Answer

Differentiate both sides with respect to x. The derivative of x^2 is 2x, and the derivative of y^2 is 2y(dy/dx). The derivative of 25 is 0. So 2x + 2y(dy/dx) = 0. Solving gives dy/dx = -x/y.

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Calculus: Derivatives

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