Differentiation of g(x) and evaluation of g′(π/6)

If g(x)=4cosx+2sinx+1, then g′(π/6)=

To find g′(π/6), we need to differentiate the function g(x) with respect to x and then evaluate the derivative at x = π/6

To find g′(π/6), we need to differentiate the function g(x) with respect to x and then evaluate the derivative at x = π/6.

Given that g(x) = 4cos(x) + 2sin(x) + 1, we can differentiate each term separately using the rules of differentiation.

The derivative of 4cos(x) is obtained by applying the chain rule. The derivative of cos(x) is -sin(x), so the derivative of 4cos(x) will be:
-4sin(x)

Similarly, the derivative of 2sin(x) is obtained by applying the chain rule. The derivative of sin(x) is cos(x), so the derivative of 2sin(x) will be:
2cos(x)

The derivative of the constant term 1 is 0, as the derivative of any constant is always 0.

Now, we can add up the derivatives of each term to find g′(x):
g′(x) = -4sin(x) + 2cos(x) + 0
= -4sin(x) + 2cos(x)

To find g′(π/6), we substitute π/6 for x in this equation:
g′(π/6) = -4sin(π/6) + 2cos(π/6)

The trigonometric values of sin(π/6) and cos(π/6) are well-known. sin(π/6) = 1/2 and cos(π/6) = √3/2 (or approximately 0.866).

Substituting these values, we get:
g′(π/6) = -4(1/2) + 2(√3/2)
= -2 + √3

Therefore, g′(π/6) is equal to -2 + √3.

More Answers: