Determine the derivative of $y(x)=\dfrac{2^x+1}{\frac{1}{2}x^2+x+1}$ in $x=0$.
Antwoord 1 correct
Correct
Antwoord 2 correct
Fout
Antwoord 3 optie
2
Antwoord 3 correct
Fout
Antwoord 4 optie
$-1$
Antwoord 4 correct
Fout
Antwoord 1 optie
$\ln(2)-2$
Antwoord 1 feedback
Correct: $y'(x)=\dfrac{2^x\ln(2)\cdot(\frac{1}{2}x^2+x+1)-(2^x+1)(x+1)}{(\frac{1}{2}x^2+x+1)^2}$.
Hence, $y'(0)=\dfrac{1\ln(2)1-2\cdot1}{1^2}=\ln(2)-2$.
Go on.
Hence, $y'(0)=\dfrac{1\ln(2)1-2\cdot1}{1^2}=\ln(2)-2$.
Go on.
Antwoord 2 feedback
Wrong: $y'(0) \neq 0$.
Try again.
Try again.
Antwoord 3 feedback
Wrong: The question is not to determine $y(0)$.
Try again.
Try again.
Antwoord 4 feedback
Wrong: $2^0 \neq 0$ and the derivative of $2^x$ is not equal to $2^x$.
See Extra explanation or Derivatives elementary functions.
See Extra explanation or Derivatives elementary functions.