derivative y=((x^2+1)/(x^2-1))^6

Lösung

ableitung von

y = (\frac{x ^{2} + 1}{x ^{2} - 1})^{6}

- \frac{24 x ( x ^{2} + 1 ) ^{5}}{( x ^{2} - 1 ) ^{7}}

Schritte zur Lösung

\frac{d}{d x} ((\frac{x ^{2} + 1}{x ^{2} - 1})^{6})

Wende die Kettenregel an:

6 (\frac{x ^{2} + 1}{x ^{2} - 1})^{5} \frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} - 1})

= 6 (\frac{x ^{2} + 1}{x ^{2} - 1})^{5} \frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} - 1})

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} - 1}) = - \frac{4 x}{( x ^{2} - 1 ) ^{2}}

= 6 (\frac{x ^{2} + 1}{x ^{2} - 1})^{5} (- \frac{4 x}{( x ^{2} - 1 ) ^{2}})

Vereinfache

6 (\frac{x ^{2} + 1}{x ^{2} - 1})^{5} (- \frac{4 x}{( x ^{2} - 1 ) ^{2}}) : - \frac{24 x ( x ^{2} + 1 ) ^{5}}{( x ^{2} - 1 ) ^{7}}

= - \frac{24 x ( x ^{2} + 1 ) ^{5}}{( x ^{2} - 1 ) ^{7}}

t an g e n t f (x) = 7 - 6 x, at x = - 1

\frac{d}{d x} (ln (tan (\frac{x}{2} + \frac{π}{4})))

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t an g e n t f (x) = \frac{5}{x}, at x = \frac{1}{4}

\int (y - x) d x