Question

find $f_{xy}(x,y)$ if $f(x,y)=e^{8xy^2}$.

$f_{xy}(x,y)=\square$

Explanation:

Step1: Find \( f_x(x,y) \)

Treat \( y \) as constant. Use chain rule: derivative of \( e^u \) is \( e^u \cdot u' \), where \( u = 8xy^2 \).
\( f_x(x,y) = e^{8xy^2} \cdot \frac{\partial}{\partial x}(8xy^2) = e^{8xy^2} \cdot 8y^2 = 8y^2 e^{8xy^2} \).

Step2: Find \( f_{xy}(x,y) \) (differentiate \( f_x \) w.r. to \( y \))

Use product rule: \( (uv)' = u'v + uv' \), where \( u = 8y^2 \), \( v = e^{8xy^2} \).

• \( u' = 16y \)
• \( v' = e^{8xy^2} \cdot \frac{\partial}{\partial y}(8xy^2) = e^{8xy^2} \cdot 16xy \)

Thus,
\( f_{xy}(x,y) = 16y \cdot e^{8xy^2} + 8y^2 \cdot 16xy e^{8xy^2} \)
Simplify:
\( = 16y e^{8xy^2} + 128xy^3 e^{8xy^2} \)
Factor:
\( = 16y e^{8xy^2}(1 + 8xy^2) \)

Answer:

\( 16y e^{8xy^2}(1 + 8xy^2) \) (or expanded as \( 16y e^{8xy^2} + 128xy^3 e^{8xy^2} \))