derivative (1/(y^2)-9/(y^4))(y+5y^3)

Lösung

ableitung von

(\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) (y + 5 y^{3})

\frac{44}{y ^{2}} + \frac{27}{y ^{4}} + 5

Schritte zur Lösung

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) (y + 5 y^{3}))

Wende die Produktregel an:

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

f = \frac{1}{y ^{2}} - \frac{9}{y ^{4}}, g = y + 5 y^{3}

= \frac{d}{d y} (\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) (y + 5 y^{3}) + \frac{d}{d y} (y + 5 y^{3}) (\frac{1}{y ^{2}} - \frac{9}{y ^{4}})

\frac{d}{d y} (\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) = - \frac{2}{y ^{3}} + \frac{36}{y ^{5}}

\frac{d}{d y} (y + 5 y^{3}) = 1 + 15 y^{2}

= (- \frac{2}{y ^{3}} + \frac{36}{y ^{5}}) (y + 5 y^{3}) + (1 + 15 y^{2}) (\frac{1}{y ^{2}} - \frac{9}{y ^{4}})

Vereinfache

(- \frac{2}{y ^{3}} + \frac{36}{y ^{5}}) (y + 5 y^{3}) + (1 + 15 y^{2}) (\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) : \frac{44}{y ^{2}} + \frac{27}{y ^{4}} + 5

= \frac{44}{y ^{2}} + \frac{27}{y ^{4}} + 5

t a y l or f (x) = ln (\frac{( 1 + x )}{( 1 - x )})

8 (ln (y)) y^{^{'}} - x^{7} y = 0

\frac{d}{d t} (cos (5 t))

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

- 6 x^{2} y + y^{^{'}} = - 10 x^{2}