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

解

\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}}

解答ステップ

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

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

であれば

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}

分離する

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}}