Let Y = 7x sin-1(2x)+^3 square root 1+4x2. Find Y'.

I don't know how to do the square root part.
I have done
y'= (7x)' (Sin^-1(2x)) + (7x)(sin^-1)'+......

y'= 7 sin^-1 + (7x)sq(2x)'/square root (1-(2x))^2 + .....

Does ^3square root mean cube root? I'll assume so. Roots are just fractional exponents, so

y = 7x arcsin(2x)+(1+4x^2)^(1/3)

recall that
d/dx arcsin(u) = 1/√(1-u^2) du/dx, so

Now, using the product rule,

y' = 7 arcsin(2x) + 14x/√(1-4x^2) + (1/3)(1+4x^2)^(-2/3) (8x)
= 7 arcsin(2x) + 14x/√(1-4x^2) + 8x/(3∛(1+4x^2)^2)

Verify this at

http://www.wolframalpha.com/input/?i=derivative+7x+arcsin%282x%29%2B%E2%88%9B%281%2B4x^2%29

If I have misinterpreted something, fix it and work the same steps. You can also fix it at wolframalpha to check your work.

To differentiate the square root part of the expression, you can use the chain rule. Let's break it down step by step:

1. Start with the expression: √(1+4x^2).

2. Let u = 1 + 4x^2. Now, rewrite the expression as: √u.

3. The derivative of √u with respect to x is given by:
(√u)' = (d√u/du) * (du/dx).

4. To find (d√u/du), you can use the power rule for differentiation:
(d√u/du) = 1/2u^(-1/2).

5. To find (du/dx), differentiate u with respect to x:
(du/dx) = d/dx(1 + 4x^2) = 8x.

6. Substitute the values back into the chain rule formula:
(√u)' = (1/2u^(-1/2)) * (8x) = 4x/u^(1/2).

Now, you can continue differentiating the rest of the expression using the chain rule. Your final expression will include all the terms you mentioned, plus the correct derivative of the square root component.