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

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

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(\partial)/(\partial α)(sin(α)cos(β))

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

cos (β) cos (α)

Solution steps

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

Treat

β

as a constant

Take the constant out:

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

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

Apply the common derivative:

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

= cos (β) cos (α)

\int \frac{1}{x ^{2} 9 - 4 x ^{2}} d x

\frac{d}{d y} (\frac{x}{x + y})

8 x - x^{2} (1.7)

x \to 0 lim (\frac{x}{- 2 x ^{5} + 15 x})

\int (1 - x) e^{x} d x

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