What is the first (cos^2(x))^' ?

The first (cos^2(x))^' is -sin(2x)

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(cos^2(x))^'

(cos^{2} (x))^{^{'}}

- sin (2 x)

Solution steps

(cos^{2} (x))^{'}

Apply the chain rule:

2 cos (x) (cos (x))^{'}

= 2 cos (x) (cos (x))^{'}

(cos (x))^{'} = - sin (x)

= 2 cos (x) (- sin (x))

Simplify

2 cos (x) (- sin (x)) : - sin (2 x)

= - sin (2 x)

(- 6, - 4), (- 4, 0)

\int_{- 1}^{2} x^{2} - x - 2 d x

\frac{\partial}{\partial x} (x ln (x))

n \to \infty lim (n^{4} + 2 n^{3} - (n^{2} + n))

\frac{8 x}{e ^{x}}