What is the derivative of 2sin^3(t) ?

The derivative of 2sin^3(t) is 6sin^2(t)cos(t)

What is the first derivative of 2sin^3(t) ?

The first derivative of 2sin^3(t) is 6sin^2(t)cos(t)

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derivative of

2 sin^{3} (t)

6 sin^{2} (t) cos (t)

Solution steps

\frac{d}{d t} (2 sin^{3} (t))

Take the constant out:

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

= 2 \frac{d}{d t} (sin^{3} (t))

Apply the chain rule:

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

= 3 (sin (t))^{2} \frac{d}{d t} (sin (t))

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

= 2 \cdot 3 (sin (t))^{2} cos (t)

Simplify

= 6 sin^{2} (t) cos (t)

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

\frac{d}{d x} (\frac{3 x + 2}{x ^{2} - 5 x + 3})

\frac{d}{d x} (cos ((8 x^{2} + 9)^{5}))

f (x) g (x)

2 x^{2} + x - 1

What is the derivative of 2sin^3(t) ?
The derivative of 2sin^3(t) is 6sin^2(t)cos(t)
What is the first derivative of 2sin^3(t) ?
The first derivative of 2sin^3(t) is 6sin^2(t)cos(t)