What is the d/(dt)(10te^{-5t}) ?

The d/(dt)(10te^{-5t}) is 10(e^{-5t}-5e^{-5t}t)

What is the first d/(dt)(10te^{-5t}) ?

The first d/(dt)(10te^{-5t}) is 10(e^{-5t}-5e^{-5t}t)

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\frac{d}{d t} (10 t e^{- 5 t})

10 (e^{- 5 t} - 5 e^{- 5 t} t)

Solution steps

\frac{d}{d t} (10 t e^{- 5 t})

Take the constant out:

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

= 10 \frac{d}{d t} (t e^{- 5 t})

Apply the Product Rule:

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

f = t, g = e^{- 5 t}

= 10 (\frac{d t}{d t} e^{- 5 t} + \frac{d}{d t} (e^{- 5 t}) t)

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

\frac{d}{d t} (e^{- 5 t}) = - 5 e^{- 5 t}

= 10 (1 \cdot e^{- 5 t} + (- 5 e^{- 5 t}) t)

Simplify

= 10 (e^{- 5 t} - 5 e^{- 5 t} t)

ln (1 - 4 y), 0

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

\int - 5 x^{3} d x

x \to 0 lim (\frac{sin ( k t )}{t})

5 x^{3}

What is the d/(dt)(10te^{-5t}) ?
The d/(dt)(10te^{-5t}) is 10(e^{-5t}-5e^{-5t}t)
What is the first d/(dt)(10te^{-5t}) ?
The first d/(dt)(10te^{-5t}) is 10(e^{-5t}-5e^{-5t}t)