What is the d/(dt)(kte^{3t}) ?

The d/(dt)(kte^{3t}) is k(e^{3t}+3e^{3t}t)

What is the first d/(dt)(kte^{3t}) ?

The first d/(dt)(kte^{3t}) is k(e^{3t}+3e^{3t}t)

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\frac{d}{d t} (k t e^{3 t})

k (e^{3 t} + 3 e^{3 t} t)

Solution steps

\frac{d}{d t} (k t e^{3 t})

Take the constant out:

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

= k \frac{d}{d t} (t e^{3 t})

Apply the Product Rule:

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

f = t, g = e^{3 t}

= k (\frac{d t}{d t} e^{3 t} + \frac{d}{d t} (e^{3 t}) t)

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

\frac{d}{d t} (e^{3 t}) = e^{3 t} \cdot 3

= k (1 \cdot e^{3 t} + e^{3 t} \cdot 3 t)

Simplify

= k (e^{3 t} + 3 e^{3 t} t)

\frac{d}{d x} (\frac{2 x}{x - 1})

\frac{d ^{2}}{d x ^{2}} (\frac{3}{4 x ^{6}})

\int sin (a x) cos (b x) d x

x \to c lim (m x + b)

2 x^{2} - 13 + 1

What is the d/(dt)(kte^{3t}) ?
The d/(dt)(kte^{3t}) is k(e^{3t}+3e^{3t}t)
What is the first d/(dt)(kte^{3t}) ?
The first d/(dt)(kte^{3t}) is k(e^{3t}+3e^{3t}t)