find the derivative of (x^2+1) 3rd root of x^2+2

If you mean (x^2+1) * (x^2+2)^(1/3) then just use the product rule and chain rule.

y = u*v^n
y' = u'v^n + u*n*v^(n-1)*v'

u = x^2 + 1
u' = 2x

v = x^2 + 2
v' = 2x

y' = 2x*cbrt(x^2+2) + 2x*(x^2+1)* 1/3 * (x^2 + 2)^(-2/3)

Feel free to simplify further. You're doing calculus; algebra should be a breeze.

To find the derivative of a function, we can use the power rule and the chain rule. In this case, we need to find the derivative of the function h(x) = (x^2 + 1) * (x^2 + 2)^(1/3).

First, let's simplify the function by expanding it:

h(x) = (x^2 + 1) * (x^2 + 2)^(1/3)
= (x^2 + 1) * [(x^2 + 2)^(1/3)]

Now, let's apply the power rule and the chain rule step by step:

1. Power rule: The derivative of x^n with respect to x is n*x^(n-1).
So, the derivative of x^2 with respect to x is d/dx (x^2) = 2x.

2. Chain rule: The derivative of f(g(x)) is f'(g(x)) * g'(x).
In our case, f(u) = u^(1/3) and g(x) = x^2 + 2.

Using the chain rule, the derivative of (x^2 + 2)^(1/3) with respect to x is:
d/dx [(x^2 + 2)^(1/3)] = (1/3)*(x^2 + 2)^(-2/3) * d/dx (x^2 + 2)

Now, let's substitute these derivatives back into our original function:

h'(x) = [(x^2 + 1) * d/dx (x^2 + 2)^(1/3)] + [(x^2 + 2)^(1/3) * d/dx (x^2 + 1)]

h'(x) = [(x^2 + 1) * (1/3)*(x^2 + 2)^(-2/3) * 2x] + [(x^2 + 2)^(1/3) * 2x]

After simplifying, we can combine like terms:

h'(x) = (2x^3 + 2x + 2x^2 + 2) / (3(x^2 + 2)^(2/3))

Therefore, the derivative of (x^2+1) * (x^2+2)^(1/3) is (2x^3 + 2x + 2x^2 + 2) / (3(x^2 + 2)^(2/3)).