Consider the function $Z(x) = z(x,y(x))$, where
$$z(x,y) = x^2y^2 \quad \text{and} \quad y(x) = e^{4x}.$$
Determine $Z'(x)$.
$$z(x,y) = x^2y^2 \quad \text{and} \quad y(x) = e^{4x}.$$
Determine $Z'(x)$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$Z'(x) = 2x(1+x)e^{8x}.$
Antwoord 2 correct
Fout
Antwoord 3 optie
$Z'(x) = 2xe^{4x}(e^{4x}+x).$
Antwoord 3 correct
Fout
Antwoord 4 optie
This derivative cannot be determined, because there is no function given for $x$.
Antwoord 4 correct
Fout
Antwoord 1 optie
$Z'(x) = 2x(1+4x)e^{8x}.$
Antwoord 1 feedback
Correct: The partial derivatives of $z(x,y)$ at $(x,y(x))$ and the derivative of $y(x)$ are:
$$
\begin{align*}
z'_x(x,y) &= y^2 \cdot 2x = 2xy^2\\
z'_x(x,y(x)) &= 2x \cdot \big(e^{4x}\big)^2 = 2xe^{8x} \\[2mm]
z'_y(x,y) &= x^2 \cdot 2y = 2x^2y\\
z'_y(x,y(x)) &= 2x^2 \cdot e^{4x} = 2x^2e^{4x}\\[2mm]
y'(x) &= e^{4x} \cdot 4 = 4e^{4x}.
\end{align*}
$$
According to the Chain rule (case 2):
$$
\begin{align*}
Z'(x) &= 2xe^{8x} + 2x^2e^{4x} \cdot 4e^{4x} = 2xe^{8x} + 8x^2e^{8x} = 2xe^{8x}(1+4x).
\end{align*}
$$
Go on.
$$
\begin{align*}
z'_x(x,y) &= y^2 \cdot 2x = 2xy^2\\
z'_x(x,y(x)) &= 2x \cdot \big(e^{4x}\big)^2 = 2xe^{8x} \\[2mm]
z'_y(x,y) &= x^2 \cdot 2y = 2x^2y\\
z'_y(x,y(x)) &= 2x^2 \cdot e^{4x} = 2x^2e^{4x}\\[2mm]
y'(x) &= e^{4x} \cdot 4 = 4e^{4x}.
\end{align*}
$$
According to the Chain rule (case 2):
$$
\begin{align*}
Z'(x) &= 2xe^{8x} + 2x^2e^{4x} \cdot 4e^{4x} = 2xe^{8x} + 8x^2e^{8x} = 2xe^{8x}(1+4x).
\end{align*}
$$
Go on.
Antwoord 2 feedback
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.
See Chain rule (case 2), Example 1 and Example 2.
Antwoord 4 feedback
Wrong: We are dealing with special case 2 of the chain rule, not case 1.
See Chain rule (case 2), Example 1 and Example 2.
See Chain rule (case 2), Example 1 and Example 2.