What is the derivative of 5e^{8t+9} ?

The derivative of 5e^{8t+9} is 40e^{8t+9}

What is the first derivative of 5e^{8t+9} ?

The first derivative of 5e^{8t+9} is 40e^{8t+9}

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derivative of

5 e^{8 t + 9}

40 e^{8 t + 9}

Solution steps

\frac{d}{d t} (5 e^{8 t + 9})

Take the constant out:

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

= 5 \frac{d}{d t} (e^{8 t + 9})

Apply the chain rule:

e^{8 t + 9} \frac{d}{d t} (8 t + 9)

= e^{8 t + 9} \frac{d}{d t} (8 t + 9)

\frac{d}{d t} (8 t + 9) = 8

= 5 e^{8 t + 9} \cdot 8

Simplify

= 40 e^{8 t + 9}

\frac{2 x ^{3} + 5}{x}

y = sin^{5} (x)

\int \frac{x - 2}{x ^{2} + 3 x} d x

n = 0 \sum \infty 2^{(3 - 2 n)}

x = 1 - y^{2}, x = y^{2} - 1

What is the derivative of 5e^{8t+9} ?
The derivative of 5e^{8t+9} is 40e^{8t+9}
What is the first derivative of 5e^{8t+9} ?
The first derivative of 5e^{8t+9} is 40e^{8t+9}