What is the d/(dt)(4e^{t/2}) ?

The d/(dt)(4e^{t/2}) is 2e^{t/2}

What is the first d/(dt)(4e^{t/2}) ?

The first d/(dt)(4e^{t/2}) is 2e^{t/2}

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\frac{d}{d t} (4 e^{\frac{t}{2}})

2 e^{\frac{t}{2}}

Solution steps

\frac{d}{d t} (4 e^{\frac{t}{2}})

Take the constant out:

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

= 4 \frac{d}{d t} (e^{\frac{t}{2}})

Apply the chain rule:

e^{\frac{t}{2}} \frac{d}{d t} (\frac{t}{2})

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

\frac{d}{d t} (\frac{t}{2}) = \frac{1}{2}

= 4 e^{\frac{t}{2}} \frac{1}{2}

Simplify

4 e^{\frac{t}{2}} \frac{1}{2} : 2 e^{\frac{t}{2}}

= 2 e^{\frac{t}{2}}

arcsin (6 x + 1)

\frac{d}{d x} (\frac{3}{x - 1})

f (x) = 6 x e^{x} csc (x)

(- 5, 7), (3, - 5)

n = 1 \sum \infty \frac{4 n}{7 n - 1}