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

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

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(\partial)/(\partial φ)(sin(θ)cos(φ))

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

- sin (θ) sin (φ)

Solution steps

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

Treat

θ

as a constant

Take the constant out:

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

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

Apply the common derivative:

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

= sin (θ) (- sin (φ))

Simplify

= - sin (θ) sin (φ)

\int 1 + 16 x^{2} d x

(x + 1) y^{^{'}} + y = x

x \to - 3 lim (5)

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

x \to - 1 lim (\frac{4 x + 3}{2 x + 1})

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