Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

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. $\boldsymbol{\frac {2x^2+\,x\,+\,2} {2\sqrt{x^2\,+\:1}} }$

[ 5-1703 - op net sinds 2.7.07-(E)-26.9.2024 ]

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

$\\ D\left ( x\sqrt{x^2+1} \right )=\left ( Dx \right )\cdot \sqrt{x^2+1}+x\cdot D\sqrt{x^2+1}\\$
$=\sqrt{x^2+1}+x\frac{1}{2\sqrt{x^2+1}}.D(x^2+1)\\ =\sqrt{x^2+1}+\frac{x^2}{\sqrt{x^2+1}} =\frac{x^2+1+x^2}{\sqrt{x^2+1}}\\ =\frac{2x^2+1}{\sqrt{x^2+1}}$

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. \(\boldsymbol{\frac {2x^2+\,x\,+\,2} {2\sqrt{x^2\,+\:1}} }\)

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. \(\boldsymbol{\frac {2x^2+\,x\,+\,2} {2\sqrt{x^2\,+\:1}} }\)