Use the rule for the deriviative of a function of a function.
Let u(v) = 6 arcsinv and v(x)= x^3
f(x) = u(v(x))
f'(x) = du/dv * dv/dx
= [6/sqrt(1-v^2)]* 2x^2
= 12 x^2/sqrt(1-x^6)]
Let u(v) = 6 arcsinv and v(x)= x^3
f(x) = u(v(x))
f'(x) = du/dv * dv/dx
= [6/sqrt(1-v^2)]* 2x^2
= 12 x^2/sqrt(1-x^6)]
Let's break down the function into the outer function and the inner function:
Outer function: f(x) = 6arcsin(u)
Inner function: u = x^3
Now, let's find the derivative of each part:
Step 1: Find the derivative of the inner function u = x^3:
Using the power rule, we can differentiate the function with respect to x:
du/dx = d/dx(x^3) = 3x^2
Step 2: Find the derivative of the outer function f(u) = 6arcsin(u):
To differentiate the arcsin function, we use the chain rule again. The derivative of arcsin(u) with respect to u is 1/sqrt(1-u^2). Therefore, we have:
df/du = 1/sqrt(1-u^2)
Step 3: Apply the chain rule:
To find f'(x), we multiply the derivatives of the outer and inner functions:
f'(x) = (df/du) * (du/dx)
= (1/sqrt(1-u^2)) * (3x^2)
Finally, substitute back the value of u:
f'(x) = (1/sqrt(1-(x^3)^2)) * (3x^2)
Therefore, the derivative of f(x) = 6arcsin(x^3) is f'(x) = (1/sqrt(1-(x^3)^2)) * (3x^2).