What is the d/(dt)(t^9e^t) ?

The d/(dt)(t^9e^t) is 9t^8e^t+e^tt^9

What is the first d/(dt)(t^9e^t) ?

The first d/(dt)(t^9e^t) is 9t^8e^t+e^tt^9

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

9 t^{8} e^{t} + e^{t} t^{9}

Solution steps

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

Apply the Product Rule:

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

f = t^{9}, g = e^{t}

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

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

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

= 9 t^{8} e^{t} + e^{t} t^{9}

4^{^{'}}

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

\int \frac{8}{x 64 - ( ln ( x ) ) ^{2}} d x

\int sec^{6} (5 x) tan (5 x) d x

(1 - x^{2})^{^{'}}