What is (\partial)/(\partial θ)((cos(θ))/(r(θ))) ?

The answer to (\partial)/(\partial θ)((cos(θ))/(r(θ))) is (-θsin(θ)-cos(θ))/(θ^2r)

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(\partial)/(\partial θ)((cos(θ))/(r(θ)))

\frac{\partial}{\partial θ} (\frac{cos ( θ )}{r ( θ )})

\frac{- θ sin ( θ ) - cos ( θ )}{θ ^{2} r}

Solution steps

\frac{\partial}{\partial θ} (\frac{cos ( θ )}{r ( θ )})

Treat

r

as a constant

Take the constant out:

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

= \frac{1}{r} \frac{\partial}{\partial θ} (\frac{cos ( θ )}{θ})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{r} \cdot \frac{\frac{\partial}{\partial θ} ( cos ( θ ) ) θ - \frac{\partial}{\partial θ} ( θ ) cos ( θ )}{θ ^{2}}

\frac{\partial}{\partial θ} (cos (θ)) = - sin (θ)

\frac{\partial}{\partial θ} (θ) = 1

= \frac{1}{r} \cdot \frac{( - sin ( θ ) ) θ - 1 \cdot cos ( θ )}{θ ^{2}}

Simplify

\frac{1}{r} \cdot \frac{( - sin ( θ ) ) θ - 1 \cdot cos ( θ )}{θ ^{2}} : \frac{- θ sin ( θ ) - cos ( θ )}{θ ^{2} r}

= \frac{- θ sin ( θ ) - cos ( θ )}{θ ^{2} r}

x \to 2 lim (\frac{0}{( x - 2 ) ^{2}})

\int 4 e^{6} d x

x \to 3 lim (x - x + 1)

\frac{\partial}{\partial y} (4 x^{3} y^{2} - 4)

\int \frac{1}{x ^{2} 6 x + 1} d x

What is (\partial)/(\partial θ)((cos(θ))/(r(θ))) ?
The answer to (\partial)/(\partial θ)((cos(θ))/(r(θ))) is (-θsin(θ)-cos(θ))/(θ^2r)