derivative of ln(y)= 1/3 ln(x^2-9)

Lösung

\frac{d}{d x} (ln (y)) = \frac{1}{3} ln (x^{2} - 9)

y = e^{\frac{1}{3} (x l n (x^{2} - 9) + 6 (\frac{1}{2} l n (\frac{x}{3} + 1) - \frac{1}{2} l n (\frac{x}{3} - 1) - \frac{x}{3})) + c_{1}}

Schritte zur Lösung

\frac{d}{d x} (ln (y)) = \frac{1}{3} ln (x^{2} - 9)

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

ln (y) = \int \frac{1}{3} ln (x^{2} - 9) d x

\int \frac{1}{3} ln (x^{2} - 9) d x = \frac{1}{3} (x ln (x^{2} - 9) + 6 (\frac{1}{2} ln (\frac{x}{3} + 1) - \frac{1}{2} ln (\frac{x}{3} - 1) - \frac{x}{3})) + c_{1}

ln (y) = \frac{1}{3} (x ln (x^{2} - 9) + 6 (\frac{1}{2} ln (\frac{x}{3} + 1) - \frac{1}{2} ln (\frac{x}{3} - 1) - \frac{x}{3})) + c_{1}

Isoliere

y : y = e^{\frac{1}{3} (x l n (x^{2} - 9) + 6 (\frac{1}{2} l n (\frac{x}{3} + 1) - \frac{1}{2} l n (\frac{x}{3} - 1) - \frac{x}{3})) + c_{1}}

y = e^{\frac{1}{3} (x l n (x^{2} - 9) + 6 (\frac{1}{2} l n (\frac{x}{3} + 1) - \frac{1}{2} l n (\frac{x}{3} - 1) - \frac{x}{3})) + c_{1}}

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

\frac{d Q}{d x} = 5 x^{4} Q

\frac{d v}{d t} = 10 + 2 v

y^{^{'}} = y^{- 9} x^{9}, y (0) = 9

\frac{d y}{d x} = - 12 x^{3} y