Consider the function $y(x) = \dfrac{2+x^2}{5-3x^3}$. Determine the derivative of this function.
Antwoord 1 correct
Correct
Antwoord 2 optie
$y'(x) = -\dfrac{2}{9x}$.
Antwoord 2 correct
Fout
Antwoord 3 optie
$y'(x) = \dfrac{-3x^4 - 18x^2 - 10x}{9x^6 - 30x^3 + 25}$.
Antwoord 3 correct
Fout
Antwoord 4 optie
None of the other answers is correct.
Antwoord 4 correct
Fout
Antwoord 1 optie
$y'(x) = \dfrac{3x^4 + 18x^2 + 10x}{9x^6 - 30x^3 + 25}$.
Antwoord 1 feedback
Correct: Write $y(x) = \dfrac{u(x)}{v(x)}$, where $u(x) = 2+x^2$ and $v(x)=5-3x^3$. By the use of the quotient rule we then find:
$$
\begin{align*}
u'(x) &= 0 + 2x = 2x\\
v'(x) &= 0 - 3\cdot 3x^2 = -9x^2\\
y'(x) &= \dfrac{u'(x)v(x) - u(x)v'(x)}{\big(v(x)\big)^2} = \dfrac{2x(5-3x^3) - (2+x^2)(-9x^2)}{\big(5-3x^3\big)^2} = \dfrac{10x -6x^4 +18x^2+9x^4}{25-30x^3+9x^6} \\
&= \dfrac{3x^4 + 18x^2 + 10x}{9x^6 - 30x^3 + 25}.
\end{align*}$$
Go on.
$$
\begin{align*}
u'(x) &= 0 + 2x = 2x\\
v'(x) &= 0 - 3\cdot 3x^2 = -9x^2\\
y'(x) &= \dfrac{u'(x)v(x) - u(x)v'(x)}{\big(v(x)\big)^2} = \dfrac{2x(5-3x^3) - (2+x^2)(-9x^2)}{\big(5-3x^3\big)^2} = \dfrac{10x -6x^4 +18x^2+9x^4}{25-30x^3+9x^6} \\
&= \dfrac{3x^4 + 18x^2 + 10x}{9x^6 - 30x^3 + 25}.
\end{align*}$$
Go on.
Antwoord 2 feedback
Wrong: You need the quotient rule to determine the derivative of $y(x)$. Note, if $y(x)=\dfrac{u(x)}{v(x)}$, then
$$ y'(x) \neq \dfrac{u'(x)}{v'(x)}.$$
See Quotient rule.
$$ y'(x) \neq \dfrac{u'(x)}{v'(x)}.$$
See Quotient rule.
Antwoord 3 feedback
Wrong: Pay attention to the order of $u'(x)v(x)$ and $u(x)v'(x)$ in the nominator of the fraction.
See Quotient rule.
See Quotient rule.
Antwoord 4 feedback
Wrong: The correct answer is among them. Maybe you have to rewrite your answer a little (work out brackets) to find the right form.
Try again.
Try again.