Consider the function $Z(x) = z(x,y(x))$, where
$$z(x,y) = e^{x^2 + y} \quad \text{and} \quad y(x) = \ln(x^2 + 5).$$
Determine $Z'(x)$.
$$z(x,y) = e^{x^2 + y} \quad \text{and} \quad y(x) = \ln(x^2 + 5).$$
Determine $Z'(x)$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$Z'(x) = e^{x^2+\ln(x^2+5)} + \dfrac{2xe^{x^2 + \ln(x^2+5)}}{x^2+5}.$
Antwoord 2 correct
Fout
Antwoord 3 optie
$Z'(x) = (2x+1)e^{x^2+\ln(x^2+5)}.$
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
$Z'(x) = 2x(x^2+6)e^{x^2}.$
Antwoord 1 feedback
Correct: The partial derivatives of $z(x,y)$ at $(x,y(x))$ and the derivative of $y(x)$ are:
$$
\begin{align*}
z'_x(x,y) &= e^{x^2+y}\cdot 2x = 2xe^{x^2+y}\\
z'_x(x,y(x)) &= 2xe^{x^2 + \ln(x^2+5)}=2xe^{x^2}e^{\ln(x^2+5)} = 2x(x^2+5)e^{x^2} \\[2mm]
z'_y(x,y) &= e^{x^2+y}\cdot 1 = e^{x^2+y}\\
z'_y(x,y(x)) &= e^{x^2+\ln(x^2+5)} = e^{x^2}e^{\ln(x^2+5)}= (x^2+5)e^{x^2}\\[2mm]
y'(x) &= \dfrac{1}{x^2+5} \cdot 2x = \dfrac{2x}{x^2+5}.
\end{align*}
$$
According to the Chain rule (case 2):
$$
\begin{align*}
Z'(x) &= 2x(x^2+5)e^{x^2} + (x^2+5)e^{x^2} \cdot \dfrac{2x}{x^2+5}= 2x(x^2+5)e^{x^2} + 2xe^{x^2} \\
&= 2xe^{x^2}(x^2+5+1) = 2x(x^2+6)e^{x^2}.
\end{align*}
$$
Go on.
$$
\begin{align*}
z'_x(x,y) &= e^{x^2+y}\cdot 2x = 2xe^{x^2+y}\\
z'_x(x,y(x)) &= 2xe^{x^2 + \ln(x^2+5)}=2xe^{x^2}e^{\ln(x^2+5)} = 2x(x^2+5)e^{x^2} \\[2mm]
z'_y(x,y) &= e^{x^2+y}\cdot 1 = e^{x^2+y}\\
z'_y(x,y(x)) &= e^{x^2+\ln(x^2+5)} = e^{x^2}e^{\ln(x^2+5)}= (x^2+5)e^{x^2}\\[2mm]
y'(x) &= \dfrac{1}{x^2+5} \cdot 2x = \dfrac{2x}{x^2+5}.
\end{align*}
$$
According to the Chain rule (case 2):
$$
\begin{align*}
Z'(x) &= 2x(x^2+5)e^{x^2} + (x^2+5)e^{x^2} \cdot \dfrac{2x}{x^2+5}= 2x(x^2+5)e^{x^2} + 2xe^{x^2} \\
&= 2xe^{x^2}(x^2+5+1) = 2x(x^2+6)e^{x^2}.
\end{align*}
$$
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 special case 1.
See Chain rule (case 2), Example 1 and Example 2.
See Chain rule (case 2), Example 1 and Example 2.