Consider the function
$$z(x,y) = \dfrac{y}{x} + e^{3y^2 + y}.$$
Determine $z'_y(x,y)$.
$$z(x,y) = \dfrac{y}{x} + e^{3y^2 + y}.$$
Determine $z'_y(x,y)$.
Antwoord 1 correct
Fout
Antwoord 2 optie
$z'_y(x,y) = \dfrac{1}{x} + e^{3y^2+y}.$
Antwoord 2 correct
Fout
Antwoord 3 optie
$z'_y(x,y) = \dfrac{-y}{x^2}.$
Antwoord 3 correct
Fout
Antwoord 4 optie
$z'_y(x,y) = \dfrac{1}{x} + (6y+1)e^{3y^2+y}.$
Antwoord 4 correct
Correct
Antwoord 1 optie
$z'_y(x,y) = \dfrac{x-y}{x^2} + (6y+1)e^{3y^2+y}.$
Antwoord 1 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 2 feedback
Antwoord 3 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 4 feedback
Correct: We have to determine the partial derivative with respect to $y$. Hence, we can consider $x$ as a constant. Then the partial derivative of $z(x,y)$ with respect to $y$ is:
$$z'_x(x,y) = \dfrac{1}{x} \cdot 1 + e^{3y^2+y} \cdot (3\cdot 2y + 1) = \dfrac{1}{x} + (6y+1)e^{3y^2+y}.$$
Here we use the sum rule, the scalar product rule and the chain rule.
Go on.
$$z'_x(x,y) = \dfrac{1}{x} \cdot 1 + e^{3y^2+y} \cdot (3\cdot 2y + 1) = \dfrac{1}{x} + (6y+1)e^{3y^2+y}.$$
Here we use the sum rule, the scalar product rule and the chain rule.
Go on.