Exercise | Wiskunde op Tilburg University

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.

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.

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.

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.