What is the d/(dt)(acos(4t)+bsin(4t)) ?

The d/(dt)(acos(4t)+bsin(4t)) is -4asin(4t)+bcos(4t)*4

What is the first d/(dt)(acos(4t)+bsin(4t)) ?

The first d/(dt)(acos(4t)+bsin(4t)) is -4asin(4t)+bcos(4t)*4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} (a cos (4 t) + b sin (4 t))

- 4 a sin (4 t) + b cos (4 t) \cdot 4

Solution steps

\frac{d}{d t} (a cos (4 t) + b sin (4 t))

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (a cos (4 t)) + \frac{d}{d t} (b sin (4 t))

\frac{d}{d t} (a cos (4 t)) = - 4 a sin (4 t)

\frac{d}{d t} (b sin (4 t)) = b cos (4 t) \cdot 4

= - 4 a sin (4 t) + b cos (4 t) \cdot 4

ln (2 cos^{4} (θ))

(x e^{x})^{^{'}}

\frac{d y}{d x} = y^{3} x e^{x^{2}}, y (0) = 1

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

What is the d/(dt)(acos(4t)+bsin(4t)) ?
The d/(dt)(acos(4t)+bsin(4t)) is -4asin(4t)+bcos(4t)*4
What is the first d/(dt)(acos(4t)+bsin(4t)) ?
The first d/(dt)(acos(4t)+bsin(4t)) is -4asin(4t)+bcos(4t)*4