(i know i've asked so many questions. i just want to make sure my answers are right for this homework! hopefully it isn't too annoying for you :) )

Find the derivative of the given function. y=(tan^-1)√(3x)
A. (1)/(√(1-3x))
B. (1)/(6√(3x(1+3x)))
C. (3)/(2(1+3x)√(3x))
D. (1)/(1+3x)

My answer was D. Although when I computed the answer and did tan^-1 as arctan, I kept getting (1)/(3x^2 +1) as my answer but it wasn't a choice. Am I doing something wrong? Again, so so sorry for asking so many questions.

since (arctan(u))' = 1/(1+u^2) u', we have

d/dx arctan √(3x)
= 1/(1+3x) (√3)/(2√(3x))
= √3 / (6x+2)√x

So, looks like C to me.

No worries at all! It's completely fine to ask questions and seek clarification.

To find the derivative of the given function y = (tan^-1)√(3x), we can use the chain rule.

Let u = √(3x), and v = tan^-1(u). We need to find dy/dx, which is dv/dx multiplied by du/dx.

First, let's find du/dx:
Using the chain rule, we have
du/dx = d/dx(√(3x))
= (1/2) * (3x)^(-1/2) * d/dx(3x)
= (1/2) * (3/2) * (x^(-1/2))
= (3/4) * (x^(-1/2))
= (3/4√x)

Next, let's find dv/du:
Using the derivative of the arctan function, we have
dv/du = d/dx(tan^-1(u))
= 1/(1 + u^2) * du/dx
= 1/(1 + u^2) * (3/4√x)
= (3/4) * (1/(1 + (3x) + 1)) * √(1/x)
= (3/4√x(1 + 3x))

Finally, we can find dy/dx by multiplying dv/du and du/dx:
dy/dx = dv/dx * du/dx
= ((3)/(4√x(1 + 3x))) * ((3)/(4√x))
= (9)/(16x(1 + 3x))

Therefore, the correct answer is not among the given choices. The answer you obtained, (1)/(3x^2 + 1), is also incorrect.

No need to apologize for asking questions. I'm here to help! If you have any more questions, feel free to ask.