Consider the function
$$z(x,y) = \dfrac{\ln(x^2 + 1)}{1+\ln(4-y)}.$$
Determine $z'_x(1,3)$.
$$z(x,y) = \dfrac{\ln(x^2 + 1)}{1+\ln(4-y)}.$$
Determine $z'_x(1,3)$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$z'_x(1,3) = 1 + \ln(2).$
Antwoord 2 correct
Fout
Antwoord 3 optie
$z'_x(1,3) = \dfrac{1}{2}.$
Antwoord 3 correct
Fout
Antwoord 4 optie
$z'_x(1,3) = \ln(2).$
Antwoord 4 correct
Fout
Antwoord 1 optie
$z'_x(1,3) = 1.$
Antwoord 1 feedback
Correct: We have to determine the partial derivative with respect to $x$. Hence, we can consider $y$, and also $\ln(4-y)$, as a constant. The partial derivative of $z(x,y)$ with respect to $x$ is:
$$z'_x(x,y) = \dfrac{\tfrac{1}{x^2+1}\cdot 2x}{1+\ln(4-y)} = \dfrac{2x}{(x^2+1)(1+\ln(4-y))}.$$
Finally, we plug $(x,y)=(1,3)$ in:
$$z'_x(1,3) = \dfrac{2\cdot 1}{(1^2+1)(1+\ln(4-3))} = \dfrac{2}{2\cdot (1+0)} = 1.$$
Go on.
$$z'_x(x,y) = \dfrac{\tfrac{1}{x^2+1}\cdot 2x}{1+\ln(4-y)} = \dfrac{2x}{(x^2+1)(1+\ln(4-y))}.$$
Finally, we plug $(x,y)=(1,3)$ in:
$$z'_x(1,3) = \dfrac{2\cdot 1}{(1^2+1)(1+\ln(4-3))} = \dfrac{2}{2\cdot (1+0)} = 1.$$
Go on.
Antwoord 2 feedback
Wrong: With respect to which variable should $z(x,y)$ be differentiated?
See Partial derivatives, Example 1, Example 2 and Example 3.
See Partial derivatives, Example 1, Example 2 and Example 3.
Antwoord 3 feedback
Antwoord 4 feedback
Wrong: With respect to which variable should $z(x,y)$ be differentiated?
See Partial derivatives, Example 1, Example 2 and Example 3.
See Partial derivatives, Example 1, Example 2 and Example 3.