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

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

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

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

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

- 2 sin (2 t)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

= - 2 sin (2 t)

\int x (3 x^{2} - 5)^{10} d x

\int x^{4} e^{5 x} d x

f (x) = \frac{3}{x + 5}

d er i v a t i v e y = \frac{3}{x ^{6}} + \frac{1}{x ^{5}} - \frac{7}{x ^{2}}

\frac{\partial}{\partial x} (x \cdot sin (4 \cdot y \cdot z^{4}))