What is the first (sin^2(t))^' ?

The first (sin^2(t))^' is sin(2t)

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(sin^2(t))^'

(sin^{2} (t))^{^{'}}

sin (2 t)

Solution steps

(sin^{2} (t))^{'}

Apply the chain rule:

2 sin (t) (sin (t))^{'}

= 2 sin (t) (sin (t))^{'}

(sin (t))^{'} = cos (t)

= 2 sin (t) cos (t)

Use the Double Angle identity:

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

= sin (2 t)

\frac{d}{d x} (- 18 sin (3 x))

x \to 0 + lim (x^{4})

x \to - 3 lim (\frac{x + 1}{x + 3})

(19, 3), (20, 3)

\int (x - \frac{2}{x})^{2} d x