What is the d/(dt)(20sin(2t-3x)) ?

The d/(dt)(20sin(2t-3x)) is 20cos(2t-3x)(2-3(dx)/(dt))

What is the first d/(dt)(20sin(2t-3x)) ?

The first d/(dt)(20sin(2t-3x)) is 20cos(2t-3x)(2-3(dx)/(dt))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} (20 sin (2 t - 3 x))

20 cos (2 t - 3 x) (2 - 3 \frac{d x}{d t})

Solution steps

\frac{d}{d t} (20 sin (2 t - 3 x))

Take the constant out:

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

= 20 \frac{d}{d t} (sin (2 t - 3 x))

Apply the chain rule:

cos (2 t - 3 x) \frac{d}{d t} (2 t - 3 x)

= cos (2 t - 3 x) \frac{d}{d t} (2 t - 3 x)

\frac{d}{d t} (2 t - 3 x) = 2 - 3 \frac{d x}{d t}

= 20 cos (2 t - 3 x) (2 - 3 \frac{d x}{d t})

y = 3 x^{7}

2 y^{2} + 1

\frac{d}{d t} (a (cos (t) + lo g_{10} (\frac{tan ( t )}{2})))

\frac{( 1 + x )}{( 1 + x ^{2} )}

\frac{x + 1}{2 x - 1}

What is the d/(dt)(20sin(2t-3x)) ?
The d/(dt)(20sin(2t-3x)) is 20cos(2t-3x)(2-3(dx)/(dt))
What is the first d/(dt)(20sin(2t-3x)) ?
The first d/(dt)(20sin(2t-3x)) is 20cos(2t-3x)(2-3(dx)/(dt))