What is the d/(dt)((s)(1/(9pi)cos(9t)-1/(3pi)sin(3t))) ?

The d/(dt)((s)(1/(9pi)cos(9t)-1/(3pi)sin(3t))) is s(-1/pi sin(9t)-1/pi cos(3t))

What is the first d/(dt)((s)(1/(9pi)cos(9t)-1/(3pi)sin(3t))) ?

The first d/(dt)((s)(1/(9pi)cos(9t)-1/(3pi)sin(3t))) is s(-1/pi sin(9t)-1/pi cos(3t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} ((s) (\frac{1}{9 π} cos (9 t) - \frac{1}{3 π} sin (3 t)))

s (- \frac{1}{π} sin (9 t) - \frac{1}{π} cos (3 t))

Solution steps

\frac{d}{d t} ((s) (\frac{1}{9 π} cos (9 t) - \frac{1}{3 π} sin (3 t)))

Take the constant out:

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

= s \frac{d}{d t} (\frac{1}{9 π} cos (9 t) - \frac{1}{3 π} sin (3 t))

Apply the Sum/Difference Rule:

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

= s (\frac{d}{d t} (\frac{1}{9 π} cos (9 t)) - \frac{d}{d t} (\frac{1}{3 π} sin (3 t)))

\frac{d}{d t} (\frac{1}{9 π} cos (9 t)) = - \frac{1}{π} sin (9 t)

\frac{d}{d t} (\frac{1}{3 π} sin (3 t)) = \frac{1}{π} cos (3 t)

= s (- \frac{1}{π} sin (9 t) - \frac{1}{π} cos (3 t))

\frac{9 x ^{2} + 4}{5 ( 3 x ^{3} + 4 x ) ^{\frac{4}{5}}}

\frac{x ( x + y + z )}{z ( x - y + z )}

(\frac{63}{32})^{2}

\frac{- 4 x ^{2} + 4 x + 4}{( x ^{2} + 1 ) ^{2}}

3098.5 (x + y - 0.1)^{2}

What is the d/(dt)((s)(1/(9pi)cos(9t)-1/(3pi)sin(3t))) ?
The d/(dt)((s)(1/(9pi)cos(9t)-1/(3pi)sin(3t))) is s(-1/pi sin(9t)-1/pi cos(3t))
What is the first d/(dt)((s)(1/(9pi)cos(9t)-1/(3pi)sin(3t))) ?
The first d/(dt)((s)(1/(9pi)cos(9t)-1/(3pi)sin(3t))) is s(-1/pi sin(9t)-1/pi cos(3t))