What is the derivative of f(t)=e^{t+9} ?

The derivative of f(t)=e^{t+9} is e^{t+9}

What is the first derivative of f(t)=e^{t+9} ?

The first derivative of f(t)=e^{t+9} is e^{t+9}

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derivative of

f (t) = e^{t + 9}

e^{t + 9}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d t} (t + 9) = 1

= e^{t + 9} \cdot 1

Simplify

= e^{t + 9}

x \to + 1 lim (\frac{4}{x - 1})

\frac{d}{d x} (e^{- 4 x} - 4 e^{- 4 x} x)

\frac{d}{d x} (ln (\frac{( x + y )}{( x - y )}))

\frac{d}{d x} (e^{x^{9}})

\frac{d}{d x} (- 5 e^{- 2.5 x^{2}} x)

What is the derivative of f(t)=e^{t+9} ?
The derivative of f(t)=e^{t+9} is e^{t+9}
What is the first derivative of f(t)=e^{t+9} ?
The first derivative of f(t)=e^{t+9} is e^{t+9}