Reelle Funktion/Ableitung/1/Aufgabe/Lösung
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Reelle Funktion/Ableitung/1/Aufgabe
Es ist
f
′
(
x
)
=
(
ln
(
x
2
+
3
)
−
x
x
2
+
2
)
′
(
1
+
sin
2
x
)
−
(
ln
(
x
2
+
3
)
−
x
x
2
+
2
)
(
1
+
sin
2
x
)
′
(
1
+
sin
2
x
)
2
=
(
2
x
x
2
+
3
−
x
2
+
2
−
x
x
x
2
+
2
)
(
1
+
sin
2
x
)
−
(
ln
(
x
2
+
3
)
−
x
x
2
+
2
)
(
2
sin
x
cos
x
)
(
1
+
sin
2
x
)
2
=
(
2
x
x
2
+
3
−
2
x
2
+
2
x
2
+
2
)
(
1
+
sin
2
x
)
−
(
ln
(
x
2
+
3
)
−
x
x
2
+
2
)
(
2
sin
x
cos
x
)
(
1
+
sin
2
x
)
2
{\displaystyle {}{\begin{aligned}f'(x)&={\frac {{\left(\ln \left(x^{2}+3\right)-x{\sqrt {x^{2}+2}}\right)}'{\left(1+\sin ^{2}x\right)}-{\left(\ln \left(x^{2}+3\right)-x{\sqrt {x^{2}+2}}\right)}{\left(1+\sin ^{2}x\right)}'}{{\left(1+\sin ^{2}x\right)}^{2}}}\\&={\frac {{\left({\frac {2x}{x^{2}+3}}-{\sqrt {x^{2}+2}}-x{\frac {x}{\sqrt {x^{2}+2}}}\right)}{\left(1+\sin ^{2}x\right)}-{\left(\ln \left(x^{2}+3\right)-x{\sqrt {x^{2}+2}}\right)}{\left(2\sin x\cos x\right)}}{{\left(1+\sin ^{2}x\right)}^{2}}}\\&={\frac {{\left({\frac {2x}{x^{2}+3}}-{\frac {2x^{2}+2}{\sqrt {x^{2}+2}}}\right)}{\left(1+\sin ^{2}x\right)}-{\left(\ln \left(x^{2}+3\right)-x{\sqrt {x^{2}+2}}\right)}{\left(2\sin x\cos x\right)}}{{\left(1+\sin ^{2}x\right)}^{2}}}\end{aligned}}}
Zur gelösten Aufgabe
