Question

Find dy over dx for y equals the product of the square root of x and the quantity 3 times x plus 1? 1) dy over dx equals the quotient of 3 and the quantity 2 times square root of x 2) dy over dx equals the quotient of the quantity 9 times x plus 1 and 2 times the square root of x 3) dy over dx equals the quotient of the quantity 9 times x minus 1 and 2 times the square root of x 4) dy over dx equals the quotient of the quantity 6 times the square root of x plus 3 times x plus 1 and 2 times the square root of x

Answer 1

The derivative (dy/dx) equals the quotient of the quantity 6 times the square root of x plus 3 times x plus 1 and 2 times the square root of x. Option 4 is answer.

Step-by-step explanation:

Given function: y = sqrt(x) * (3x + 1)

Using the product rule (uv)' = u'v + uv', where u = sqrt(x) and v = 3x + 1:

Calculate the derivative of sqrt(x): d/dx(sqrt(x)) = 1 / (2 * sqrt(x))

Calculate the derivative of (3x + 1): d/dx(3x + 1) = 3

Substitute these into the product rule formula:

dy/dx = (1 / (2 * sqrt(x))) * (3x + 1) + sqrt(x) * 3

Simplify the expression:

dy/dx = (3x + 1) / (2 * sqrt(x)) + 3 * sqrt(x)

Combine the terms with a common denominator:

dy/dx = (3x + 1 + 6x * sqrt(x)) / (2 * sqrt(x))

Factor out a common factor of x in the numerator:

dy/dx = [x * (6 * sqrt(x) + 3) + 1] / (2 * sqrt(x))

The final expression matches the correct option 4.