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

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

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

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

ρ cos (θ) cos (φ)

Solution steps

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

Treat

ρ, θ

as constants

Take the constant out:

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

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

Apply the common derivative:

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

= ρ cos (θ) cos (φ)

y^{^{''}} - 4 y^{^{'}} + 5 y = 2020 e^{- t}

5 2 x^{3} + 8 x (2.2)

(- 7.1) (- 7.9)

(x - 2)^{5}

n \to \infty lim (5 n e^{- n})

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