What is the (40cos(20t-pi/2))^' ?

The (40cos(20t-pi/2))^' is 800cos(20t)

What is the first (40cos(20t-pi/2))^' ?

The first (40cos(20t-pi/2))^' is 800cos(20t)

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(40 cos (20 t - \frac{π}{2}))^{^{'}}

800 cos (20 t)

Solution steps

(40 cos (20 t - \frac{π}{2}))^{'}

Simplify

40 cos (20 t - \frac{π}{2}) : 40 sin (20 t)

= (40 sin (20 t))^{^{'}}

Take the constant out:

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

= 40 (sin (20 t))^{^{'}}

Apply the chain rule:

cos (20 t) (20 t)^{'}

= cos (20 t) (20 t)^{'}

(20 t)^{'} = 20

= 40 cos (20 t) \cdot 20

Simplify

= 800 cos (20 t)

\int t e^{- 7 t} d t

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

\int \frac{1}{4 + 3 x} d x

\frac{d y}{d x} = e^{y + 2 x}

y^{^{''}} + y^{^{'}} = 2 e^{x} + 14

What is the (40cos(20t-pi/2))^' ?
The (40cos(20t-pi/2))^' is 800cos(20t)
What is the first (40cos(20t-pi/2))^' ?
The first (40cos(20t-pi/2))^' is 800cos(20t)