Consider the function z(x,y)=2^{xy^2}+\dfrac{e^x}{\sqrt{x}}+\;^4\!\log (y^3-7). Determine z''_{xy}.
Antwoord 1 correct
Correct
Antwoord 2 optie
z''_{xy}(x,y)=2y\cdot 2^{xy^2} \cdot (1+y^2x)
Antwoord 2 correct
Fout
Antwoord 3 optie
z''_{xy}(x,y)=2^{xy^2}\cdot (\textrm{ln}(2))^2
Antwoord 3 correct
Fout
Antwoord 4 optie
z''_{xy}(x,y)=2y^2x\cdot 2^{xy^2}\cdot \textrm{ln}(2)
Antwoord 4 correct
Fout
Antwoord 1 optie
z''_{xy}(x,y)=2\cdot \textrm{ln}(2)y\cdot 2^{xy^2} \cdot (1+y^2x\cdot \textrm{ln}(2))
Antwoord 1 feedback
Correct: We rewrite z(x,y)=2^{xy^2}+f(x)+g(y), with f(x)=\dfrac{e^x}{\sqrt{x}} and g(y)=\;^4\!\log (y^3-7).
Then z'_x(x,y)=y^2\cdot 2^{xy^2}\cdot \textrm{ln}(2)+f'(x).
Hence, \begin{align*} z''_{xy} (x,y)&=z''_{yx}(x,y)\\ &=2y\cdot 2^{xy^2}\cdot \textrm{ln}(2)+y^2\cdot 2^{xy^2}\cdot 2yx \cdot (\textrm{ln}(2))^2\\ &=2\cdot \textrm{ln}(2)y\cdot 2^{xy^2} \cdot (1+y^2x\cdot \textrm{ln}(2)) \end{align*}
Go on.
Then z'_x(x,y)=y^2\cdot 2^{xy^2}\cdot \textrm{ln}(2)+f'(x).
Hence, \begin{align*} z''_{xy} (x,y)&=z''_{yx}(x,y)\\ &=2y\cdot 2^{xy^2}\cdot \textrm{ln}(2)+y^2\cdot 2^{xy^2}\cdot 2yx \cdot (\textrm{ln}(2))^2\\ &=2\cdot \textrm{ln}(2)y\cdot 2^{xy^2} \cdot (1+y^2x\cdot \textrm{ln}(2)) \end{align*}
Go on.
Antwoord 2 feedback
Antwoord 3 feedback
Antwoord 4 feedback