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

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

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(\partial)/(\partial θ)(sin(θ)cos(φ))

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

cos (φ) cos (θ)

Solution steps

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

Treat

φ

as a constant

Take the constant out:

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

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

Apply the common derivative:

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

= cos (φ) cos (θ)

\int 4 x - 6 x y^{2} - 2 d x

t^{2} y^{^{''}} + 19 t y^{^{'}} + 130 y = 0

\frac{\partial}{\partial x} (e^{- xx} sin (3 x + 1))

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

n = 1 \sum \infty \frac{( n + 6 )}{2}

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