Consider the function
$$ z(x,y) =xy^2 + x^3y.$$
Use the property of the partial derivatives to determine by how much the variable $x$ approximately has to change when $y$ increases by 0.4 such that the function value remains the same with respect to $z(1,2)$.
$\Delta x \approx -0.8.$
$\Delta x \approx 2.$
Cannot be determined, because we only know the approximation of the change of the function value.
$\Delta x \approx -0.2.$
Correct: The change in the function value is approximately equal to
$$\Delta z \approx z'_x(x_0,y_0) \Delta x + z'_y(x_0,y_0) \Delta y,$$
which means that the change in $x$ is approximately
$$\Delta x \approx \dfrac{\Delta z - z'_y(x_0,y_0) \Delta y}{z'_x(x_0,y_0)}.$$
It is given that
$$(x_0,y_0) = (1,2), \qquad \Delta y = 0.4 \qquad \text{and} \qquad \Delta z = 0.$$
The partial derivatives at $(1,2)$ are
$$
\begin{align*}
z'_x(x,y) &= y^2 + 3x^2y &&& z'_x(1,2) &= 2^2 + 3\cdot1^2\cdot2 = 10,\\
z'_y(x,y) &=2xy + x^3 &&& z'_y(1,2) &= 2\cdot1\cdot2 + 1^3 = 5.
\end{align*}
$$
Hence, the needed approximate change in $x$ is
$$\Delta x \approx \dfrac{\Delta z - z'_y(1,2) \Delta y}{z'_x(1,2)} =\dfrac{0- 5 \cdot 0.4}{10}=-0.2.$$
Go on.
Wrong: What is it you have to determine? And what is given?
See Property partial derivatives and the examples.
Wrong: What is it you have to determine? And what is given?
See Property partial derivatives and the examples.
Wrong: $\Delta x$ can certainly be determined.
See Property partial derivatives and the examples.