without any simplification, find the derivative of
h(x) Arcsin^4 (3x-1) / (3)^1/2 (square root)
I end up with h'(x) =
12Arcsin^3 (3x-1)sqr(3) / 3 (squr(1-(3-x)^2))
I'm just not sure if I have to do the product of both Arcsin^4 (3x-1) and then do the quotient rule with sqr(3), or do it the way I just did, considering Arcsin^4 (3x-1) as one and do the quotient rule right away
Thank you
You are correct, except that in the denominator, that should be (3x-1)^2, not (3-x)^2
If you simplify the radicand by expanding and folding in the sqrt(3) from the numerator, you end up with
4Arcsin^3 (3x-1) / sqrt(2x-3x^2)
There is no quotient here. sqrt(3) is just a constant, and you just have u^4 where u = arcsin(v) and v = 3x-1; use the regular power rule, and the chain rule twice.
To find the derivative of the given function h(x) = Arcsin^4 (3x-1) / (√3), we'll use the chain rule. The chain rule states that if we have a composite function of the form f(g(x)), then its derivative is given by f'(g(x)) * g'(x).
Let's break down the steps:
Step 1: Identify the outer function and the inner function.
The outer function is Arcsin^4(x), and the inner function is (3x-1)/√3.
Step 2: Find the derivative of the outer function.
To find the derivative of Arcsin^4(x), we'll use the general power rule. The derivative of x^n, where n is a constant, is given by n*x^(n-1). Applying this rule, we get:
d/dx [Arcsin^4(x)] = 4 * Arcsin^3(x) * (d/dx) [Arcsin(x)].
Step 3: Find the derivative of the inner function.
The derivative of (3x-1)/√3 can be found using the quotient rule, which states that if we have a function of the form f(x)/g(x), the derivative is [f'(x)*g(x) - f(x)*g'(x)] / [g(x)]^2. In this case, f(x) = (3x-1) and g(x) = √3. Let's calculate:
(d/dx) [(3x-1)/√3] = [3*√3 - (3x-1)*(1/2*√3)] / (√3)^2
= [3√3 - (3x-1)/(2√3)] / 3
= [6√3 - 3x + 1] / (6√3).
Step 4: Combine the derivative of the outer function with the derivative of the inner function.
Using the chain rule, we multiply the derivative of the outer function (4 * Arcsin^3(x)) with the derivative of the inner function ([6√3 - 3x + 1] / (6√3)). Thus, we obtain:
h'(x) = [4 * Arcsin^3 (3x-1)] * ([6√3 - 3x + 1] / (6√3)).
After simplification, this can be written as:
h'(x) = [12 * Arcsin^3 (3x-1) * √3] / (6√3 - 3x + 1).
Therefore, the derivative of h(x) = Arcsin^4 (3x-1) / (√3) is:
h'(x) = [12 * Arcsin^3 (3x-1) * √3] / (6√3 - 3x + 1).