derivative of (sin^2(x)/(1+cos^2(x)))

Lösung

\frac{d}{d x} (\frac{sin ^{2} ( x )}{1 + cos ^{2} ( x )})

\frac{2 sin ( 2 x )}{( 1 + cos ^{2} ( x ) ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{sin ^{2} ( x )}{1 + cos ^{2} ( x )})

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (sin^{2} (x)) = sin (2 x)

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

= \frac{sin ( 2 x ) ( 1 + cos ^{2} ( x ) ) - ( - sin ( 2 x ) ) sin ^{2} ( x )}{( 1 + cos ^{2} ( x ) ) ^{2}}

Vereinfache

\frac{sin ( 2 x ) ( 1 + cos ^{2} ( x ) ) - ( - sin ( 2 x ) ) sin ^{2} ( x )}{( 1 + cos ^{2} ( x ) ) ^{2}} : \frac{2 sin ( 2 x )}{( 1 + cos ^{2} ( x ) ) ^{2}}

= \frac{2 sin ( 2 x )}{( 1 + cos ^{2} ( x ) ) ^{2}}

\int \frac{1}{( 4 x + 5 ) ^{2}} d x

t an g e n t f (x) = 7 x + 1, at x = 9

\int \frac{2 x + 1}{x ^{6} + 18 x ^{4} + 81 x ^{2}} d x

(\frac{lo g _{10} ( x )}{x})^{^{'}}

t an g e n t \frac{6}{5 x + 3}