Calculus: Derivatives
Finding and interpreting rates of change
Calculus: Derivatives
Finding and interpreting rates of change
Calculus - Grade 9-12
- 1
Use the limit definition of the derivative to find f'(x) for f(x) = x^2.
Expand (x + h)^2 before simplifying the fraction.
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. - 2
Find the derivative of f(x) = 7x^5 - 3x^2 + 9x - 4.
Use the power rule on each term separately.
The derivative is f'(x) = 35x^4 - 6x + 9. The constant term -4 has derivative 0. - 3
Find dy/dx if y = 4x^-3 + 6sqrt(x). Write your answer using exponents or radicals.
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). - 4
At x = 2, find the slope of the tangent line to the curve f(x) = x^3 - 4x + 1.
The derivative gives the slope of the tangent line at a point.
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. - 5
Find the equation of the tangent line to f(x) = x^2 + 3x at x = 1.
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. - 6
Find the derivative of g(x) = (2x^3 - x)(5x^2 + 1).
Use (uv)' = u'v + uv'.
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. - 7
Find the derivative of h(x) = (x^2 + 1)/(x - 3).
Use (u/v)' = (vu' - uv')/v^2.
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. - 8
Find the derivative of f(x) = (3x - 2)^5.
Differentiate the outside function, then multiply by the derivative of the inside function.
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. - 9
Find the derivative of y = sin(x) + 4cos(x) - 2tan(x).
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). - 10
Find the derivative of y = e^x ln(x), where x is greater than 0.
Both factors change with x, so use the product rule.
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. - 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.
Velocity is the rate of change of position with respect to time.
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. - 12
For the car in the previous problem, find the acceleration function.
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. - 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).
Estimate the derivative by finding the slope of the tangent line.
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. - 14
A function has derivative f'(x) = 3x^2 - 12x + 9. Find the x-values where the graph of f has horizontal tangent lines.
A horizontal tangent line has slope 0.
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. - 15
Use implicit differentiation to find dy/dx for x^2 + y^2 = 25.
Remember that y is a function of x, so the derivative of y^2 is 2y(dy/dx).
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.