Find f'(x) for f(x) = sin3(3x2).
look at it as:
f(x) = (sin(3x^2))^3
f ' (x) = 3(sin(3x^2))^2 cos(3x^2) (6x)
= 18x sin^2 (3x^2) cos(3x^2)
or
= 9x sin(6x^2) cos(3x^2), using property sin 2A = 2sinAcosA
Sorry it is sin^3(3x^2)
Hmm. I see an extra sin, not cos:
9x sin(6x^2) sin(3x^2)
To find the derivative of f(x) = sin^3(3x^2), we can use the chain rule.
Step 1: Identify the inner function and the outer function.
In this case, the outer function is sin^3(u), where u is the inner function 3x^2.
Step 2: Find the derivative of the outer function.
The derivative of sin^3(u) with respect to u is 3sin^2(u)cos(u). Note that we use the derivative of sin^3(u) = 3sin^2(u)cos(u).
Step 3: Find the derivative of the inner function.
The derivative of 3x^2 with respect to x is 6x.
Step 4: Combine the derivatives using the chain rule.
Multiplying the derivative of the outer function (3sin^2(u)cos(u)) with the derivative of the inner function (6x), we get:
f'(x) = 6x * 3sin^2(3x^2) * cos(3x^2)
Thus, the derivative of f(x) = sin^3(3x^2) is f'(x) = 6x * 3sin^2(3x^2) * cos(3x^2).