y = √3x + 2 cos(x), 0 ≤ x < 2π
Find a derivative y'.
To find the derivative of y = √(3x) + 2cos(x), we can use the rules of differentiation.
First, let's differentiate each term separately:
1. Differentiating √(3x):
The derivative of √(3x) can be found using the chain rule. Let u = 3x, then √(3x) = u^(1/2). Now, we can find the derivative of u^(1/2) with respect to x using the power rule, which states that the derivative of x^n is nx^(n-1).
So, the derivative of √(3x) with respect to x is (1/2)u^(-1/2) * du/dx.
Now, we need to find du/dx. Since u = 3x, the derivative of u with respect to x is simply 3.
Substituting these values back into the equation, we get:
dy/dx = (1/2)(3x)^(-1/2) * 3
2. Differentiating 2cos(x):
The derivative of cos(x) is -sin(x). Since we have a coefficient of 2 in front of cos(x), we need to multiply the derivative by 2.
So, the derivative of 2cos(x) with respect to x is -2sin(x).
Now, we can add these derivatives together to find the derivative of y:
dy/dx = (1/2)(3x)^(-1/2) * 3 + (-2sin(x))
Simplifying this expression further, we get:
dy/dx = (3/2√(3x)) - 2sin(x)
Therefore, the derivative of y with respect to x, y', is (3/2√(3x)) - 2sin(x).