Consider the function $Z(x) = z(x,y(x))$, where
$$z(x,y) = \ln(x^2 + y^3) \quad \text{and} \quad y(x) = 10-x^2.$$
Determine $Z'(3)$.
$Z'(3) = -\dfrac{6}{5}.$
$Z'(3) = -\dfrac{1}{2}.$
$Z'(3) = \dfrac{9}{10}.$
$Z'(3)=-\dfrac{13}{3}.$
Consider the function $Z(x) = z(x,y(x))$, where
$$z(x,y) = \ln(x^2 + y^3) \quad \text{and} \quad y(x) = 10-x^2.$$
Determine $Z'(3)$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$Z'(3) = -\dfrac{1}{2}.$
Antwoord 2 correct
Fout
Antwoord 3 optie
$Z'(3) = \dfrac{9}{10}.$
Antwoord 3 correct
Fout
Antwoord 4 optie
$Z'(3)=-\dfrac{13}{3}.$
Antwoord 4 correct
Fout
Antwoord 1 optie
$Z'(3) = -\dfrac{6}{5}.$
Antwoord 1 feedback
Correct: The partial derivatives of $z(x,y)$ at $(x,y(x))$ and the derivative of $y(x)$ are:
$$
\begin{align*}
z'_x(x,y) &= \dfrac{1}{x^2 + y^3} \cdot 2x = \dfrac{2x}{x^2 + y^3}\\
z'_x(x,y(x)) &= \dfrac{2x}{x^2 + (10-x^2)^3}\\[2mm]
z'_y(x,y) &=\dfrac{1}{x^2 + y^3} \cdot 3y^2 = \dfrac{3y^2}{x^2 + y^3}\\
z'_y(x,y(x)) &= \dfrac{3(10-x^2)^2}{x^2 + (10-x^2)^3}\\[2mm]
y'(x) &= -2x.
\end{align*}
$$
According to the Chain rule (case 2):
$$
\begin{align*}
Z'(x) &= \dfrac{2x}{x^2 + (10-x^2)^3} + \dfrac{3(10-x^2)^2}{x^2 + (10-x^2)^3} \cdot (-2x) = \dfrac{2x}{x^2 + (10-x^2)^3} + \dfrac{-6x(10-x^2)^2}{x^2 + (10-x^2)^3} \\
&= \dfrac{2x -6x(10-x^2)^2}{x^2 + (10-x^2)^3}.
\end{align*}
$$
Finally, we plug in $x=3$:
$$
\begin{align*}
Z'(3) &= \dfrac{2\cdot 3 -6\cdot 3(10-3^2)^2}{3^2 + (10-3^2)^3} = \dfrac{-12}{10} = -\dfrac{6}{5}.
\end{align*}
$$
Go on.
Antwoord 2 feedback
Wrong: $z'_x(x,y) \neq \dfrac{1}{x^2+y^3}$ and $z'_y(x,y) \neq \dfrac{1}{x^2+y^3}$.

See Chain rule.
Antwoord 3 feedback
Wrong: Do not forget to apply the chain rule: special case 2.

See Chain rule (case 2), Example 1 and Example 2.
Antwoord 4 feedback
Wrong: It is given that $x=3$, not that $y=3$.

See Chain rule (case 2), Example 1 and Example 2.