What is the derivative of 1+cos(θ) ?

The derivative of 1+cos(θ) is -sin(θ)

What is the first derivative of 1+cos(θ) ?

The first derivative of 1+cos(θ) is -sin(θ)

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derivative of

1 + cos (θ)

- sin (θ)

Solution steps

\frac{d}{d θ} (1 + cos (θ))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d θ} (1) + \frac{d}{d θ} (cos (θ))

\frac{d}{d θ} (1) = 0

\frac{d}{d θ} (cos (θ)) = - sin (θ)

= 0 - sin (θ)

Simplify

= - sin (θ)

\frac{\partial}{\partial x} (3 x^{7} y^{8} + 8 x^{5} y^{6})

\frac{d}{d x} (\frac{x ^{\frac{2}{3}}}{5 + x + x ^{4}})

\int 4 x + \frac{1}{4 x} d x

\int 5 x e^{- 2 x} d x

\frac{2}{s + 3}

What is the derivative of 1+cos(θ) ?
The derivative of 1+cos(θ) is -sin(θ)
What is the first derivative of 1+cos(θ) ?
The first derivative of 1+cos(θ) is -sin(θ)