d/(dt)(({z}(t))ln(sqrt(1+t)+(1+sqrt(t))^2))

Lösung

\frac{d}{d t} ((z (t)) ln (1 + t + (1 + t)^{2}))

ln (1 + t + (1 + t)^{2}) \frac{d}{d t} (z (t)) + \frac{( t + 2 ( 1 + t ) t + 1 ) z ( t )}{2 t ( 1 + t + ( 1 + t ) ^{2} ) t + 1}

Schritte zur Lösung

\frac{d}{d t} ((z (t)) ln (1 + t + (1 + t)^{2}))

Wende die Produktregel an:

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

f = z (t), g = ln (1 + t + (1 + t)^{2})

= \frac{d}{d t} (z (t)) ln (1 + t + (1 + t)^{2}) + \frac{d}{d t} (ln (1 + t + (1 + t)^{2})) z (t)

\frac{d}{d t} (ln (1 + t + (1 + t)^{2})) = \frac{t + 2 ( 1 + t ) t + 1}{2 t ( 1 + t + ( 1 + t ) ^{2} ) t + 1}

= \frac{d}{d t} (z (t)) ln (1 + t + (1 + t)^{2}) + \frac{t + 2 ( 1 + t ) t + 1}{2 t ( 1 + t + ( 1 + t ) ^{2} ) t + 1} z (t)

Multipliziere Brüche:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= ln (1 + t + (1 + t)^{2}) \frac{d}{d t} (z (t)) + \frac{( t + 2 ( 1 + t ) t + 1 ) z ( t )}{2 t ( 1 + t + ( 1 + t ) ^{2} ) t + 1}

e x p an d \frac{g + k}{( 1 + g ( g + k ) )}

e x p an d (- (2 x^{4} + 5 y^{5}))^{7}

\frac{d}{d s} ((k) (s^{2} + 2 s + 4 + s^{2} k + 5 s k - 9 k))

e x p an d (\frac{1 + 1}{2 x})^{8}

e x p an d \frac{( 2 c t ^{2} + 3 ) t}{2}