What is the d/(dθ)(2cos(θ)+2sin(θ)) ?

The d/(dθ)(2cos(θ)+2sin(θ)) is -2sin(θ)+2cos(θ)

What is the first d/(dθ)(2cos(θ)+2sin(θ)) ?

The first d/(dθ)(2cos(θ)+2sin(θ)) is -2sin(θ)+2cos(θ)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d θ} (2 cos (θ) + 2 sin (θ))

- 2 sin (θ) + 2 cos (θ)

Solution steps

\frac{d}{d θ} (2 cos (θ) + 2 sin (θ))

Apply the Sum/Difference Rule:

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

= \frac{d}{d θ} (2 cos (θ)) + \frac{d}{d θ} (2 sin (θ))

\frac{d}{d θ} (2 cos (θ)) = - 2 sin (θ)

\frac{d}{d θ} (2 sin (θ)) = 2 cos (θ)

= - 2 sin (θ) + 2 cos (θ)

3 (1 + x^{2}) \frac{d y}{d x} = 2 x y (y^{3} - 1)

e^{- 2 s} \cdot \frac{1}{s - 3}

\int \frac{1}{( x ^{2} - 25 ) ^{\frac{1}{2}}} d x

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

x \to 0 lim (2 x e^{2 x} + k x)

What is the d/(dθ)(2cos(θ)+2sin(θ)) ?
The d/(dθ)(2cos(θ)+2sin(θ)) is -2sin(θ)+2cos(θ)
What is the first d/(dθ)(2cos(θ)+2sin(θ)) ?
The first d/(dθ)(2cos(θ)+2sin(θ)) is -2sin(θ)+2cos(θ)