Consider the function $f(t) = z(x(t),y(t))$, where
$$z(x,y) = \ln(x^4 + y^3), \quad x(t) = 2t+1 \quad \text{and} \quad y(t) = t^2+5.$$
Determine $f'(-2)$.
$$z(x,y) = \ln(x^4 + y^3), \quad x(t) = 2t+1 \quad \text{and} \quad y(t) = t^2+5.$$
Determine $f'(-2)$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$f'(-2) = -\dfrac{1}{405}.$
Antwoord 2 correct
Fout
Antwoord 3 optie
$f'(-2) =\dfrac{1}{6}.$
Antwoord 3 correct
Fout
Antwoord 4 optie
$f'(-2) = -14.$
Antwoord 4 correct
Fout
Antwoord 1 optie
$f'(-2)=-\dfrac{22}{15}.$
Antwoord 1 feedback
Correct: Note that we have to determine the derivative of $f(t)$ in $t=-2$; hence, it holds that
$x(-2) = 2\cdot(-2)+1=-3$ and $y(-2) = (-2)^2 + 5 = 9$. The partial derivatives of $z(x,y)$ at $(x(t),y(t))=(-3,9)$, the derivative of $x(t)$ in $t=-2$ and the derivative of $y(t)$ in $t=-2$ are:
$$
\begin{align*}
z'_x(x,y) &= \dfrac{1}{x^4+y^3} \cdot 4x^3 = \dfrac{4x^3}{x^4+y^3}\\
z'_x(-3,9) &= \dfrac{4(-3)^3}{(-3)^4+9^3} = \dfrac{-108}{810} = -\dfrac{2}{15}\\[2mm]
z'_y(x,y) &= \dfrac{1}{x^4+y^3} \cdot 3y^2 = \dfrac{3y^2}{x^4+y^3}\\
z'_y(-3,9) &= \dfrac{3\cdot 9^2}{(-3)^4+9^3} = \dfrac{243}{810} = \dfrac{3}{10}\\[2mm]
x'(t) &= 2\\
x'(-2) &= 2\\[1mm]
y'(t) &= 2t\\
y'(-2) & = 2\cdot (-2) = -4.
\end{align*}
$$
According to the Chain rule (case 1):
$$
f'(-2) = z'_x(-3,9)x'(-2) + z'_y(-3,9)y'(-2) = -\dfrac{2}{15}\cdot 2 + \dfrac{3}{10} \cdot (-4) = \dfrac{-4}{15} - \dfrac{12}{10} = -\dfrac{22}{15}.
$$
Go on.
$x(-2) = 2\cdot(-2)+1=-3$ and $y(-2) = (-2)^2 + 5 = 9$. The partial derivatives of $z(x,y)$ at $(x(t),y(t))=(-3,9)$, the derivative of $x(t)$ in $t=-2$ and the derivative of $y(t)$ in $t=-2$ are:
$$
\begin{align*}
z'_x(x,y) &= \dfrac{1}{x^4+y^3} \cdot 4x^3 = \dfrac{4x^3}{x^4+y^3}\\
z'_x(-3,9) &= \dfrac{4(-3)^3}{(-3)^4+9^3} = \dfrac{-108}{810} = -\dfrac{2}{15}\\[2mm]
z'_y(x,y) &= \dfrac{1}{x^4+y^3} \cdot 3y^2 = \dfrac{3y^2}{x^4+y^3}\\
z'_y(-3,9) &= \dfrac{3\cdot 9^2}{(-3)^4+9^3} = \dfrac{243}{810} = \dfrac{3}{10}\\[2mm]
x'(t) &= 2\\
x'(-2) &= 2\\[1mm]
y'(t) &= 2t\\
y'(-2) & = 2\cdot (-2) = -4.
\end{align*}
$$
According to the Chain rule (case 1):
$$
f'(-2) = z'_x(-3,9)x'(-2) + z'_y(-3,9)y'(-2) = -\dfrac{2}{15}\cdot 2 + \dfrac{3}{10} \cdot (-4) = \dfrac{-4}{15} - \dfrac{12}{10} = -\dfrac{22}{15}.
$$
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: $x \neq-2$ and $y\neq-2$.
Try again.
Try again.