De afgeleide van het product \(\boldsymbol{x\sqrt{x^2+1} }\)
is gelijk aan
A.  \(\boldsymbol{\frac {x} {\sqrt{x^2\,+\:1}} }\)
B.  \(\boldsymbol{\frac {2x^2+\,1} {\sqrt{x^2\,+\:1}} }\)
C.  \(\boldsymbol{\frac {x^2+\,x\,+\,1} {\sqrt{x^2\,+\:1}} }\)
D.  \(\boldsymbol{\frac {4x^2+\,5} {4\sqrt{x^2\,+\:1}} }\)
E.  \(\boldsymbol{\frac {3x^2+\,1} {2\sqrt{x^2\,+\:1}} }\)
F.  \(\boldsymbol{\frac {2x^2+\,x\,+\,2} {2\sqrt{x^2\,+\:1}} }\)
A    B    C    D    E    F 

[ 5-1703 - op net sinds 2.7.07-(E)-2.11.2023 ]

Translation in   E N G L I S H

The derivative
of the product
\(\boldsymbol{x\sqrt{x^2+1} }\)
is equal to
A.   \(\boldsymbol{\frac {x} {\sqrt{x^2\,+\:1}} }\)
B.   \(\boldsymbol{\frac {2x^2+\,1} {\sqrt{x^2\,+\:1}} }\)
C.   \(\boldsymbol{\frac {x^2+\,x\,+\,1} {\sqrt{x^2\,+\:1}} }\)
D.   \(\boldsymbol{\frac {4x^2+\,5} {4\sqrt{x^2\,+\:1}} }\)
E.   \(\boldsymbol{\frac {3x^2+\,1} {2\sqrt{x^2\,+\:1}} }\)
F.   \(\boldsymbol{\frac {2x^2+\,x\,+\,2} {2\sqrt{x^2\,+\:1}} }\)

Oplossing - Solution


gricha