Answer:
y'(x) = 3x²(x² + 1)^(1/3) + (2/3)x⁴(x² + 1)^(-2/3)
Explanation:
To find the first derivative y' of the function y = x³(x² + 1)^(1/3), you will need to apply the product rule. The product rule states that if you have a function of the form u(x) * v(x), then the derivative (u * v)' is given by u'v + uv'.
In this case:
u(x) = x³
v(x) = (x² + 1)^(1/3)
Now, let's find the derivatives of u(x) and v(x):
u'(x) = 3x² (using the power rule for differentiation)
v'(x) = (1/3)(x² + 1)^(-2/3) * 2x (using the chain rule and the power rule)
Now, we can apply the product rule:
y'(x) = u'v + uv'
y'(x) = (3x²)((x² + 1)^(1/3)) + (x³)((1/3)(x² + 1)^(-2/3) * 2x)
Now, you can simplify this expression further if you like:
y'(x) = 3x²(x² + 1)^(1/3) + (2/3)x⁴(x² + 1)^(-2/3)