What is the d/(dt)(6sin^3(t)) ?

The d/(dt)(6sin^3(t)) is 18sin^2(t)cos(t)

What is the first d/(dt)(6sin^3(t)) ?

The first d/(dt)(6sin^3(t)) is 18sin^2(t)cos(t)

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\frac{d}{d t} (6 sin^{3} (t))

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

Solution steps

\frac{d}{d t} (6 sin^{3} (t))

Take the constant out:

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

= 6 \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)

= 6 \cdot 3 (sin (t))^{2} cos (t)

Simplify

= 18 sin^{2} (t) cos (t)

\frac{d ^{2}}{d x ^{2}} (\frac{5}{6 t ^{4}})

\frac{d y}{d x} = \frac{( x + y ) ^{2}}{2 x ^{2}}

\frac{d}{d x} (x^{2} - 4 x y + y^{2})

x^{2} y^{^{'}} + x y = 9

\frac{d}{d x} (ln (x) (x - 1))

What is the d/(dt)(6sin^3(t)) ?
The d/(dt)(6sin^3(t)) is 18sin^2(t)cos(t)
What is the first d/(dt)(6sin^3(t)) ?
The first d/(dt)(6sin^3(t)) is 18sin^2(t)cos(t)