What is (\partial)/(\partial θ)(acos(θ)) ?

The answer to (\partial)/(\partial θ)(acos(θ)) is -asin(θ)

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(\partial)/(\partial θ)(acos(θ))

\frac{\partial}{\partial θ} (a cos (θ))

- a sin (θ)

Solution steps

\frac{\partial}{\partial θ} (a cos (θ))

Take the constant out:

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

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

Apply the common derivative:

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

= a (- sin (θ))

Simplify

= - a sin (θ)

\frac{\partial}{\partial x} (5 x^{2} y + 6 x y^{3})

y^{^{'}} = 9 y^{2} sin (x)

f (x) = 3 x^{3} + 5 x^{2} + 5 x, at x = - 1

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

y = 2 x^{2}, y = 2 x + 6, - 2, 2

What is (\partial)/(\partial θ)(acos(θ)) ?
The answer to (\partial)/(\partial θ)(acos(θ)) is -asin(θ)