Calculus Challenge

Find the derivative of y = x^2 + 3x - 2 using implicit differentiation.

• dy/dx = 2x + 3

• dy/dx = 3x - 2

• dy/dx = 2x - 3

• dy/dx = x^2 + 3x - 2

Differentiate the function y = (2x^3 + 5x^2) / (x^2 + 1) using the quotient rule.

• (6x^2 + 10x - 2x^3 + 5x^2) / (x^2 + 1)^2

• (6x^2 + 10x - 2x^3 - 5x^2) / (x^2 - 1)^2

• (6x^2 + 10x - 2x^3 - 5x^2) / (x^2 + 1)^2

• (6x^2 + 10x - 2x^3 - 5x^2) / (x^2 + 1)

Apply the chain rule to find the derivative of y = sin(3x^2).

• y' = sin(3x^2) * 6x

• y' = cos(3x^2) * 3x

• y' = sin(6x)

• y' = cos(3x^2) * 6x

Calculate the derivative of y = ln(x^2 + 1).

• 1/(x^2 + 1)

• 2x/(x^2 - 1)

• x^2/(2x)

• 2x/(x^2 + 1)

Find the derivative of y = e^(2x + 1).

• dy/dx = e^(2x + 1) / 2

• dy/dx = e^(2x + 1) + 2

• dy/dx = e^(2x + 1) * 2

• dy/dx = e^(2x + 1) - 2

Using the quotient rule, differentiate y = (3x^2 - 2x + 1) / (x^3 + 1).

• (3x^4 - 2x^2 + 1) / (x^3 + 1)^2

• (-6x^4 - 2x^2 + 3) / (x^3 + 1)^2

• (6x^4 + 2x^2 - 3) / (x^3 - 1)

• (-6x^4 - 2x^2 + 3) / (x^3 + 1)

Apply the chain rule to find the derivative of y = cos(2x^3).

• -6x^2 * cos(2x^3)

• -6x^2 * sin(2x^3)

• 4x^2 * sin(2x^3)

• 6x^2 * sin(3x^2)

Calculate the derivative of y = ln(2x + 1) - ln(x - 1).

• y' = (2/(2x + 1)) - (1/(x - 1))

• y' = (2/(2x - 1)) - (1/(x - 1))

• y' = (2/(2x + 1)) - (1/(x + 1))

• y' = (2/(2x + 1)) + (1/(x - 1))

Find the derivative of y = e^(3x^2 + 2x).

• dy/dx = (6x + 2)e^(3x)

• dy/dx = (3x^2 + 2x)e^(3x^2 + 2x)

• dy/dx = (6x + 2)e^(3x^2 + 2x)

• dy/dx = (3x^2 + 2x)e^(6x)

Differentiate the function y = (x^3 + 2x^2) / (x^2 - 1) using the quotient rule.

• The derivative of y = (x^3 + 2x^2) / (x^2 - 1) is (x^3 + 2x^2) / (x^2 - 1)^2 + 2x / (x^2 - 1)

• The derivative of y = (x^3 + 2x^2) / (x^2 - 1) is (3x^2 + 4x) / (2x - 1)^2 + 2x / (x^2 - 1)

• The derivative of y = (x^3 + 2x^2) / (x^2 - 1) is (3x^2 + 4x) / (x^2 - 1)^2 + 4x / (x^2 - 1)

• The derivative of y = (x^3 + 2x^2) / (x^2 - 1) is (2x^3 - 2x) / (x^2 - 1)^2 + 4x / (x^2 - 1)