Consider the function $f(t) = z(x(t),y(t))$, where
$$z(x,y) = \dfrac{x}{y}, \quad x(t) = e^{2t} \quad \text{and} \quad y(t) = \ln(t^2+2).$$
Determine $Z'(t)$.
$$z(x,y) = \dfrac{x}{y}, \quad x(t) = e^{2t} \quad \text{and} \quad y(t) = \ln(t^2+2).$$
Determine $Z'(t)$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$Z'(t) = \dfrac{e^{2t}}{\ln(t^2+2)} + \dfrac{-e^{2t}}{(t^2+2)\big(\ln(t^2+2)\big)^2}.$
Antwoord 2 correct
Fout
Antwoord 3 optie
$Z'(t) =\dfrac{1}{\ln(t^2+2)} + \dfrac{e^{2t}}{\big(\ln(t^2+2)\big)^2}.$
Antwoord 3 correct
Fout
Antwoord 4 optie
$Z'(t) = \dfrac{1}{\ln(t^2+2)} + \dfrac{-2te^{2t}}{(t^2+2)\big(\ln(t^2+2)\big)^2}.$
Antwoord 4 correct
Fout
Antwoord 1 optie
$Z'(t) = \dfrac{2e^{2t}}{\ln(t^2+2)} + \dfrac{-2te^{2t}}{(t^2+2)\big(\ln(t^2+2)\big)^2}.$
Antwoord 1 feedback
Correct: The partial derivatives of $z(x,y)$ at $(x(t),y(t))$, the derivative of $x(t)$ and the derivative of $y(t)$ are:
$$
\begin{align*}
z'_x(x,y) &= \tfrac{1}{y} \cdot 1 = \tfrac{1}{y}\\
z'_x(x(t),y(t)) &= \tfrac{1}{\ln(t^2+2)}\\[2mm]
z'_y(x,y) &= x \cdot (-1) \cdot y^{-2} = \tfrac{-x}{y^2}\\
z'_y(x(t),y(t)) &= \tfrac{-e^{2t}}{\big(\ln(t^2+2)\big)^2}\\[2mm]
x'(t) &= e^{2t} \cdot 2 = 2e^{2t}\\[1mm]
y'(t) &= \tfrac{1}{t^2+2} \cdot 2t = \tfrac{2t}{t^2+2}.
\end{align*}
$$
According to Chain rule (case 1):
$$
Z'(t) = \tfrac{1}{\ln(t^2+2)} \cdot 2e^{2t}+ \tfrac{-e^{2t}}{\big(\ln(t^2+2)\big)^2} \cdot \tfrac{2t}{t^2+2} = \tfrac{2e^{2t}}{\ln(t^2+2)} + \tfrac{-2te^{2t}}{(t^2+2)\big(\ln(t^2+2)\big)^2}.
$$
Go on.
$$
\begin{align*}
z'_x(x,y) &= \tfrac{1}{y} \cdot 1 = \tfrac{1}{y}\\
z'_x(x(t),y(t)) &= \tfrac{1}{\ln(t^2+2)}\\[2mm]
z'_y(x,y) &= x \cdot (-1) \cdot y^{-2} = \tfrac{-x}{y^2}\\
z'_y(x(t),y(t)) &= \tfrac{-e^{2t}}{\big(\ln(t^2+2)\big)^2}\\[2mm]
x'(t) &= e^{2t} \cdot 2 = 2e^{2t}\\[1mm]
y'(t) &= \tfrac{1}{t^2+2} \cdot 2t = \tfrac{2t}{t^2+2}.
\end{align*}
$$
According to Chain rule (case 1):
$$
Z'(t) = \tfrac{1}{\ln(t^2+2)} \cdot 2e^{2t}+ \tfrac{-e^{2t}}{\big(\ln(t^2+2)\big)^2} \cdot \tfrac{2t}{t^2+2} = \tfrac{2e^{2t}}{\ln(t^2+2)} + \tfrac{-2te^{2t}}{(t^2+2)\big(\ln(t^2+2)\big)^2}.
$$
Go on.
Antwoord 2 feedback
Antwoord 3 feedback
Wrong: Do not forget to apply the chain rule: special case 1.
See Chain rule (case 1), Example 1 and Example 2.
See Chain rule (case 1), Example 1 and Example 2.
Antwoord 4 feedback
Wrong: We are dealing with special case 1, not with special case 2..
See Chain rule (case 1), Example 1 and Example 2.
See Chain rule (case 1), Example 1 and Example 2.