What is the d/(dt)(e^{a(e^t-1)}) ?

The d/(dt)(e^{a(e^t-1)}) is ae^{a(e^t-1)+t}

What is the first d/(dt)(e^{a(e^t-1)}) ?

The first d/(dt)(e^{a(e^t-1)}) is ae^{a(e^t-1)+t}

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\frac{d}{d t} (e^{a (e^{t} - 1)})

a e^{a (e^{t} - 1) + t}

Solution steps

\frac{d}{d t} (e^{a (e^{t} - 1)})

Apply the chain rule:

e^{a (e^{t} - 1)} \frac{d}{d t} (a (e^{t} - 1))

= e^{a (e^{t} - 1)} \frac{d}{d t} (a (e^{t} - 1))

\frac{d}{d t} (a (e^{t} - 1)) = a e^{t}

= e^{a (e^{t} - 1)} a e^{t}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e^{a (e^{t} - 1)} e^{t} = e^{a (e^{t} - 1) + t}

= a e^{a (e^{t} - 1) + t}

y^{^{''}} + 3 y^{^{'}} + 2 y = 7 cos (x) - sin (x)

\frac{\partial}{\partial z} (- z^{2})

\int \frac{6 x + 5}{( 9 x ^{2} + 15 x ) ^{5}} d x

\int_{4}^{8} t^{3} ln (5 t) d t

What is the d/(dt)(e^{a(e^t-1)}) ?
The d/(dt)(e^{a(e^t-1)}) is ae^{a(e^t-1)+t}
What is the first d/(dt)(e^{a(e^t-1)}) ?
The first d/(dt)(e^{a(e^t-1)}) is ae^{a(e^t-1)+t}