Find the derivative of 4sinx(6cosx+9sinx) at x=¦Ð/6.
To find the derivative of the given function, we can use the product rule and the chain rule. The product rule states that if we have two functions, u(x) and v(x), the derivative of their product is given by:
(d/dx)(u(x) * v(x)) = u'(x) * v(x) + u(x) * v'(x)
In this case, u(x) = 4sin(x) and v(x) = (6cos(x) + 9sin(x)). Applying the product rule, we have:
(d/dx)(4sin(x) * (6cos(x) + 9sin(x))) = (4cos(x) * (6cos(x) + 9sin(x))) + (4sin(x) * (-6sin(x) + 9cos(x)))
Next, we can simplify this expression by using the distributive property:
= (24cos^2(x) + 36sin(x)cos(x)) + (-24sin^2(x) + 36cos(x)sin(x))
Now, we can simplify further by combining like terms:
= 24cos^2(x) - 24sin^2(x) + 36sin(x)cos(x) + 36sin(x)cos(x)
Since cos^2(x) - sin^2(x) = cos(2x), we can rewrite the expression as:
= 24cos(2x) + 72sin(x)cos(x)
Now, we can substitute x = π/6 into the expression to find the derivative at x = π/6:
= 24cos(2(π/6)) + 72sin(π/6)cos(π/6)
= 24cos(π/3) + 72sin(π/6)cos(π/6)
= 24 * (1/2) + 72 * (1/2) * (sqrt(3)/2)
= 12 + 36 * (sqrt(3)/4)
= 12 + 9sqrt(3)
Therefore, the derivative of 4sin(x)(6cos(x) + 9sin(x)) at x = π/6 is 12 + 9sqrt(3).