What is the derivative of 2sin(t)cos(t) ?

The derivative of 2sin(t)cos(t) is cos(2t)*2

What is the first derivative of 2sin(t)cos(t) ?

The first derivative of 2sin(t)cos(t) is cos(2t)*2

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derivative of

2 sin (t) cos (t)

cos (2 t) \cdot 2

Solution steps

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

Simplify

2 sin (t) cos (t) : sin (2 t)

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

Apply the chain rule:

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

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

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

= cos (2 t) \cdot 2

y^{^{''}} - 4 y^{^{'}} + 5 = 0

f (x) = \frac{5}{2 5 x + 6}, at a = 2

y = \frac{x ^{3}}{9}, 0 \leq x \leq 2

t y^{^{'}} + 7 y = t^{2} - t + 1, y (1) = \frac{1}{7}

\int tan^{2} (d) x d x

What is the derivative of 2sin(t)cos(t) ?
The derivative of 2sin(t)cos(t) is cos(2t)*2
What is the first derivative of 2sin(t)cos(t) ?
The first derivative of 2sin(t)cos(t) is cos(2t)*2