(Twitter)
0
\[f(x)=-\cot(x)\cdot\frac1{\sin(x)}\\ \overset{\text{Produktregel}}\implies f'(x)=-\left( \cot'(x)\cdot\frac1{\sin(x)}-\cot(x)\left(\frac1{\sin(x)}\right)'\right)=-\left(-\frac1{\sin^3(x)}-\frac{\cos^2(x)}{\sin^3(x)}\right)=\frac{1+\cos^2(x)}{\sin^3(x)}\]
Diese Antwort melden
Link
geantwortet
ikaros
Student, Punkte: 35
Student, Punkte: 35
Also f = -cos(x), g = sin^2(x) =>
f''(x) = ((-cos(x))' * sin^2(x) -(-cos(x)) * (sin^2(x))')/(sin^2(x))^2 = ((sin(x)*sin^2(x)+cos(x)*2sin(x)*cos(x)))/sin^4(x)
─ kallemann 26.06.2020 um 15:54