What is the d/(dt)(3.5sin(5t)) ?

The d/(dt)(3.5sin(5t)) is 17.5cos(5t)

What is the first d/(dt)(3.5sin(5t)) ?

The first d/(dt)(3.5sin(5t)) is 17.5cos(5t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} (3.5 sin (5 t))

17.5 cos (5 t)

Solution steps

\frac{d}{d t} (3.5 sin (5 t))

Take the constant out:

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

= 3.5 \frac{d}{d t} (sin (5 t))

Apply the chain rule:

cos (5 t) \frac{d}{d t} (5 t)

= cos (5 t) \frac{d}{d t} (5 t)

\frac{d}{d t} (5 t) = 5

= 3.5 cos (5 t) \cdot 5

Simplify

= 17.5 cos (5 t)

(\frac{1}{x ^{3}})^{^{'}}

y = (3 x)^{5 x}

y = 9 x, y = \frac{3}{4} x, y = 70 - x^{2}

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

(1 + x)^{\frac{1}{4}}

What is the d/(dt)(3.5sin(5t)) ?
The d/(dt)(3.5sin(5t)) is 17.5cos(5t)
What is the first d/(dt)(3.5sin(5t)) ?
The first d/(dt)(3.5sin(5t)) is 17.5cos(5t)