What is the d/(dt)(3+3cos(t^3)-4sin(t^3)) ?

The d/(dt)(3+3cos(t^3)-4sin(t^3)) is -9t^2sin(t^3)-12t^2cos(t^3)

What is the first d/(dt)(3+3cos(t^3)-4sin(t^3)) ?

The first d/(dt)(3+3cos(t^3)-4sin(t^3)) is -9t^2sin(t^3)-12t^2cos(t^3)

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\frac{d}{d t} (3 + 3 cos (t^{3}) - 4 sin (t^{3}))

- 9 t^{2} sin (t^{3}) - 12 t^{2} cos (t^{3})

Solution steps

\frac{d}{d t} (3 + 3 cos (t^{3}) - 4 sin (t^{3}))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (3) + \frac{d}{d t} (3 cos (t^{3})) - \frac{d}{d t} (4 sin (t^{3}))

\frac{d}{d t} (3) = 0

\frac{d}{d t} (3 cos (t^{3})) = - 9 t^{2} sin (t^{3})

\frac{d}{d t} (4 sin (t^{3})) = 12 t^{2} cos (t^{3})

= 0 - 9 t^{2} sin (t^{3}) - 12 t^{2} cos (t^{3})

Simplify

= - 9 t^{2} sin (t^{3}) - 12 t^{2} cos (t^{3})

x^{- 10}

y^{^{'}} + 5 y = 2 e^{(6 \cdot x)}

\frac{\partial}{\partial z} (x^{2} yz - x y z^{3})

\frac{y ^{4}}{8}

\int \frac{1}{( x + 1 ) x ^{2} + 2 x} d x

What is the d/(dt)(3+3cos(t^3)-4sin(t^3)) ?
The d/(dt)(3+3cos(t^3)-4sin(t^3)) is -9t^2sin(t^3)-12t^2cos(t^3)
What is the first d/(dt)(3+3cos(t^3)-4sin(t^3)) ?
The first d/(dt)(3+3cos(t^3)-4sin(t^3)) is -9t^2sin(t^3)-12t^2cos(t^3)