Consider the function
$$ z(x,y) = x + e^{xy-6}.$$
Use the property of partial derivatives to approximate the change in function value with respect to $z(2,3)$ if $y$ increases with $\tfrac{1}{5}$, while $x$ decreases with $\tfrac{1}{4}$.
$\Delta z \approx -\tfrac{3}{5}.$
$\Delta z \approx \tfrac{3}{10}.$
$\Delta z \approx \tfrac{7}{5}.$
Cannot be determined, because the input variable $x$ is decreasing, which cannot be expressed in $\Delta x$.
Consider the function
$$ z(x,y) = x + e^{xy-6}.$$
Use the property of partial derivatives to approximate the change in function value with respect to $z(2,3)$ if $y$ increases with $\tfrac{1}{5}$, while $x$ decreases with $\tfrac{1}{4}$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$\Delta z \approx \tfrac{3}{10}.$
Antwoord 2 correct
Fout
Antwoord 3 optie
$\Delta z \approx \tfrac{7}{5}.$
Antwoord 3 correct
Fout
Antwoord 4 optie
Cannot be determined, because the input variable $x$ is decreasing, which cannot be expressed in $\Delta x$.
Antwoord 4 correct
Fout
Antwoord 1 optie
$\Delta z \approx -\tfrac{3}{5}.$
Antwoord 1 feedback
Correct: The change in function value is approximately
$$\Delta z \approx z'_x(x_0,y_0) \Delta x + z'_y(x_0,y_0) \Delta y.$$
It is given that
$$(x_0,y_0) = (2,3), \qquad \Delta x = -\tfrac{1}{4} \qquad \text{and} \qquad \Delta y = \tfrac{1}{5}.$$
The partial derivatives at $(2,3)$ are
$$
\begin{align*}
z'_x(x,y) &= 1+e^{xy-6}\cdot y = 1 + ye^{xy-6} &&& z'_x(2,3) &= 1 + 3e^{2\cdot3-6} = 4,\\
z'_y(x,y) &= e^{xy-6}\cdot x = xe^{xy-6} &&& z'_y(2,3) &= 2e^{2\cdot3-6} = 2.
\end{align*}
$$
The change in the function value is approximately equal to
$$\Delta z \approx z'_x(2,3)\Delta x + z'_y(2,3)\Delta y = 4 \cdot (-\tfrac{1}{4}) + 2 \cdot \tfrac{1}{5} = -1 + \tfrac{2}{5} = -\tfrac{3}{5}.$$

Go on.
Antwoord 2 feedback
Wrong: What is the change in $x$? And what is the change in $y$?

See Property partial derivatives, Example 1 and Example 2.
Antwoord 3 feedback
Wrong: What is the change in $x$?

See Property partial derivatives, Example 1 and Example 2.
Antwoord 4 feedback
Wrong: A decrease of the input variable corresponds to a negative value of $\Delta x$.

Possibly see Property partial derivatives, Example 1 and Example 2, and try again.