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