What is (\partial)/(\partial φ)(ρsin(φ)sin(θ)) ?

The answer to (\partial)/(\partial φ)(ρsin(φ)sin(θ)) is ρsin(θ)cos(φ)

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(\partial)/(\partial φ)(ρsin(φ)sin(θ))

\frac{\partial}{\partial φ} (ρ sin (φ) sin (θ))

ρ sin (θ) cos (φ)

Solution steps

\frac{\partial}{\partial φ} (ρ sin (φ) sin (θ))

Treat

ρ, θ

as constants

Take the constant out:

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

= ρ sin (θ) \frac{\partial}{\partial φ} (sin (φ))

Apply the common derivative:

\frac{\partial}{\partial φ} (sin (φ)) = cos (φ)

= ρ sin (θ) cos (φ)

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{e ^{2 x}}{x ^{2}}

x \to a lim (sin (x))

\int_{6}^{8} \frac{1}{x ^{2} - 1} d x

x \to 0 lim (x sec (x) csc (x))

\frac{d y}{d x} + y + 3 sin (x) = 0

What is (\partial)/(\partial φ)(ρsin(φ)sin(θ)) ?
The answer to (\partial)/(\partial φ)(ρsin(φ)sin(θ)) is ρsin(θ)cos(φ)