What is the d/(dt)((x)(cte^{-b/(2m)t})) ?

The d/(dt)((x)(cte^{-b/(2m)t})) is cx(-(be^{-\frac{bt)/(2m)}t}{2m}+e^{-(bt)/(2m)})

What is the first d/(dt)((x)(cte^{-b/(2m)t})) ?

The first d/(dt)((x)(cte^{-b/(2m)t})) is cx(-(be^{-\frac{bt)/(2m)}t}{2m}+e^{-(bt)/(2m)})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} ((x) (c t e^{- \frac{b}{2 m} t}))

c x (- \frac{b e ^{- \frac{b t}{2 m}} t}{2 m} + e^{- \frac{b t}{2 m}})

Solution steps

\frac{d}{d t} ((x) (c t e^{- \frac{b}{2 m} t}))

Simplify

(x) (c t e^{- \frac{b}{2 m} t}) : c e^{- \frac{b t}{2 m}} x t

= \frac{d}{d t} (c e^{- \frac{b t}{2 m}} x t)

Treat

c, m, b, x

as a constant

Take the constant out:

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

= c x \frac{d}{d t} (e^{- \frac{b t}{2 m}} t)

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{- \frac{b t}{2 m}}, g = t

= c x (\frac{d}{d t} (e^{- \frac{b t}{2 m}}) t + \frac{d t}{d t} e^{- \frac{b t}{2 m}})

\frac{d}{d t} (e^{- \frac{b t}{2 m}}) = - \frac{b e ^{- \frac{b t}{2 m}}}{2 m}

\frac{d t}{d t} = 1

= c x ((- \frac{b e ^{- \frac{b t}{2 m}}}{2 m}) t + 1 \cdot e^{- \frac{b t}{2 m}})

Simplify

c x ((- \frac{b e ^{- \frac{b t}{2 m}}}{2 m}) t + 1 \cdot e^{- \frac{b t}{2 m}}) : c x (- \frac{b e ^{- \frac{b t}{2 m}} t}{2 m} + e^{- \frac{b t}{2 m}})

= c x (- \frac{b e ^{- \frac{b t}{2 m}} t}{2 m} + e^{- \frac{b t}{2 m}})

2.18 \cdot 1 0^{- 18} (3^{2}) (\frac{1}{4 ^{2}} - \frac{1}{6 ^{2}})

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

e^{- l n (x)} x

e^{- 2} \cdot 2^{0} \cdot \frac{1}{0 !} \cdot 0.125

(- 3 e^{- 2 (t - x)} + 10 e^{- t + x}) \cdot e^{- 2 x}

What is the d/(dt)((x)(cte^{-b/(2m)t})) ?
The d/(dt)((x)(cte^{-b/(2m)t})) is cx(-(be^{-\frac{bt)/(2m)}t}{2m}+e^{-(bt)/(2m)})
What is the first d/(dt)((x)(cte^{-b/(2m)t})) ?
The first d/(dt)((x)(cte^{-b/(2m)t})) is cx(-(be^{-\frac{bt)/(2m)}t}{2m}+e^{-(bt)/(2m)})