What is the d/(dθ)(6sin(θ)-8cos(θ)) ?

The d/(dθ)(6sin(θ)-8cos(θ)) is 6cos(θ)+8sin(θ)

What is the first d/(dθ)(6sin(θ)-8cos(θ)) ?

The first d/(dθ)(6sin(θ)-8cos(θ)) is 6cos(θ)+8sin(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} (6 sin (θ) - 8 cos (θ))

6 cos (θ) + 8 sin (θ)

Solution steps

\frac{d}{d θ} (6 sin (θ) - 8 cos (θ))

Apply the Sum/Difference Rule:

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

= \frac{d}{d θ} (6 sin (θ)) - \frac{d}{d θ} (8 cos (θ))

\frac{d}{d θ} (6 sin (θ)) = 6 cos (θ)

\frac{d}{d θ} (8 cos (θ)) = - 8 sin (θ)

= 6 cos (θ) - (- 8 sin (θ))

Simplify

= 6 cos (θ) + 8 sin (θ)

\int 2 x^{- 2} d x

ln (4 z + 1)

x \to 1 lim (3 x^{3} - 4 x^{2} + x + 5)

\frac{d}{d x} (sin (x) + \frac{8}{7} cot (x))

\int_{- 1}^{0} (2 x + 3)^{2} d x

What is the d/(dθ)(6sin(θ)-8cos(θ)) ?
The d/(dθ)(6sin(θ)-8cos(θ)) is 6cos(θ)+8sin(θ)
What is the first d/(dθ)(6sin(θ)-8cos(θ)) ?
The first d/(dθ)(6sin(θ)-8cos(θ)) is 6cos(θ)+8sin(θ)