What is (\partial)/(\partial β)(sin(α)cos(β)) ?

The answer to (\partial)/(\partial β)(sin(α)cos(β)) is -sin(α)sin(β)

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(\partial)/(\partial β)(sin(α)cos(β))

\frac{\partial}{\partial β} (sin (α) cos (β))

- sin (α) sin (β)

Solution steps

\frac{\partial}{\partial β} (sin (α) cos (β))

Treat

α

as a constant

Take the constant out:

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

= sin (α) \frac{\partial}{\partial β} (cos (β))

Apply the common derivative:

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

= sin (α) (- sin (β))

Simplify

= - sin (α) sin (β)

\frac{\partial}{\partial y} (x^{y^{z}})

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

\frac{\partial}{\partial x} (4 x^{0.5} + 2 x^{0.25})

\int 2 x e^{4 x} d x

y^{^{'}} + 3 (tan (3 x)) y = - 10 cos (3 x)

What is (\partial)/(\partial β)(sin(α)cos(β)) ?
The answer to (\partial)/(\partial β)(sin(α)cos(β)) is -sin(α)sin(β)