derivative f(x)=(c^8-x^8)/(c^8+x^8)

Lösung

ableitung von

f (x) = \frac{c ^{8} - x ^{8}}{c ^{8} + x ^{8}}

- \frac{16 c ^{8} x ^{7}}{( c ^{8} + x ^{8} ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{c ^{8} - x ^{8}}{c ^{8} + x ^{8}})

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( c ^{8} - x ^{8} ) ( c ^{8} + x ^{8} ) - \frac{d}{d x} ( c ^{8} + x ^{8} ) ( c ^{8} - x ^{8} )}{( c ^{8} + x ^{8} ) ^{2}}

\frac{d}{d x} (c^{8} - x^{8}) = - 8 x^{7}

\frac{d}{d x} (c^{8} + x^{8}) = 8 x^{7}

= \frac{( - 8 x ^{7} ) ( c ^{8} + x ^{8} ) - 8 x ^{7} ( c ^{8} - x ^{8} )}{( c ^{8} + x ^{8} ) ^{2}}

Vereinfache

\frac{( - 8 x ^{7} ) ( c ^{8} + x ^{8} ) - 8 x ^{7} ( c ^{8} - x ^{8} )}{( c ^{8} + x ^{8} ) ^{2}} : - \frac{16 c ^{8} x ^{7}}{( c ^{8} + x ^{8} ) ^{2}}

= - \frac{16 c ^{8} x ^{7}}{( c ^{8} + x ^{8} ) ^{2}}

d er i v a t i v e e^{2 x + 3}

\int 30 - 10 t d t

\int z z - 1 d z

\int e^{0.3 x} d x

y^{^{''}} - 2 y^{^{'}} + 50 y = 0, y (0) = \frac{3}{2}, y^{^{'}} (0) = - 9