derivative e^{t/9}cos(t/9)

Lösung

ableitung von

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

\frac{1}{9} e^{\frac{t}{9}} cos (\frac{t}{9}) - \frac{1}{9} e^{\frac{t}{9}} sin (\frac{t}{9})

Schritte zur Lösung

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

Wende die Produktregel an:

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

f = e^{\frac{t}{9}}, g = cos (\frac{t}{9})

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

\frac{d}{d t} (e^{\frac{t}{9}}) = e^{\frac{t}{9}} \frac{1}{9}

\frac{d}{d t} (cos (\frac{t}{9})) = - sin (\frac{t}{9}) \frac{1}{9}

= e^{\frac{t}{9}} \frac{1}{9} cos (\frac{t}{9}) + (- sin (\frac{t}{9}) \frac{1}{9}) e^{\frac{t}{9}}

Vereinfache

= \frac{1}{9} e^{\frac{t}{9}} cos (\frac{t}{9}) - \frac{1}{9} e^{\frac{t}{9}} sin (\frac{t}{9})

\int \frac{8 x + 1}{x ^{2} + 1} d x

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y^{^{''}} + y = 9 sin (x)

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\int 3 x - 5 d x