What is the first ((x^2+1)^2)^' ?

The first ((x^2+1)^2)^' is 4x(x^2+1)

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((x^2+1)^2)^'

((x^{2} + 1)^{2})^{^{'}}

4 x (x^{2} + 1)

Solution steps

((x^{2} + 1)^{2})^{'}

Apply the chain rule:

2 (x^{2} + 1) (x^{2} + 1)^{'}

= 2 (x^{2} + 1) (x^{2} + 1)^{'}

(x^{2} + 1)^{'} = 2 x

= 2 (x^{2} + 1) \cdot 2 x

Simplify

= 4 x (x^{2} + 1)

y^{^{''}} + 4 y^{^{'}} + 4 y = sec (t) tan (t)

\int_{0}^{1} (x^{2} - x) d x

3000 (14 - x) (x - 5)

f (x) = 6 x^{2} + 8 x

\frac{x}{2 - tan ( x )}