What is the d/(dθ)(4-θ^2sin(θ)) ?

The d/(dθ)(4-θ^2sin(θ)) is -θ^2cos(θ)-2θsin(θ)

What is the first d/(dθ)(4-θ^2sin(θ)) ?

The first d/(dθ)(4-θ^2sin(θ)) is -θ^2cos(θ)-2θsin(θ)

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\frac{d}{d θ} (4 - θ^{2} sin (θ))

- θ^{2} cos (θ) - 2 θ sin (θ)

Solution steps

\frac{d}{d θ} (4 - θ^{2} sin (θ))

Apply the Sum/Difference Rule:

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

= \frac{d}{d θ} (4) - \frac{d}{d θ} (θ^{2} sin (θ))

\frac{d}{d θ} (4) = 0

\frac{d}{d θ} (θ^{2} sin (θ)) = 2 θ sin (θ) + cos (θ) θ^{2}

= 0 - (2 θ sin (θ) + cos (θ) θ^{2})

Simplify

= - θ^{2} cos (θ) - 2 θ sin (θ)

\frac{\partial}{\partial x} (\frac{1}{2 x + 3 y})

y = 3 - 2 x + 4 x^{3}, (- 1, 1)

\int 1 + cos (x) d x

x^{2} \frac{d y}{d x} - 2 x^{3} e^{- \frac{1}{x}} = y

\int sin (5 x) sin (4 x) d x

What is the d/(dθ)(4-θ^2sin(θ)) ?
The d/(dθ)(4-θ^2sin(θ)) is -θ^2cos(θ)-2θsin(θ)
What is the first d/(dθ)(4-θ^2sin(θ)) ?
The first d/(dθ)(4-θ^2sin(θ)) is -θ^2cos(θ)-2θsin(θ)