What is the d/(dt)(2sin^2(t)) ?

The d/(dt)(2sin^2(t)) is 2sin(2t)

What is the first d/(dt)(2sin^2(t)) ?

The first d/(dt)(2sin^2(t)) is 2sin(2t)

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\frac{d}{d t} (2 sin^{2} (t))

2 sin (2 t)

Solution steps

\frac{d}{d t} (2 sin^{2} (t))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d t} (sin^{2} (t))

Apply the chain rule:

2 sin (t) \frac{d}{d t} (sin (t))

= 2 sin (t) \frac{d}{d t} (sin (t))

\frac{d}{d t} (sin (t)) = cos (t)

= 2 \cdot 2 sin (t) cos (t)

Simplify

2 \cdot 2 sin (t) cos (t) : 2 sin (2 t)

= 2 sin (2 t)

x \to 0 lim (x^{2} + 2 x)

\int x^{- 0.5} d x

\frac{d}{d x} (0.0225 x e^{- 0.0467 x})

\int \frac{1}{x ^{2} + 10 x + 24} d x

\int \frac{1}{x ^{\frac{1}{5}}} d x