this is similar to section 6.1 problem 34:\nfor ( f(x,y)=2x^{3}y^{3}+6e^{xy} ):\n(a) determine ( f_{xx}(x,y)…

Explanation:

Step1: Find (f_x(x,y))

Differentiate (f(x,y)=2x^{3}y^{3}+6e^{xy}) with respect to (x) using the power rule ((x^n)^\prime = nx^{n - 1}) and the chain rule ((e^{u})^\prime=e^{u}\cdot u^\prime) (where (u = xy)). [ \begin{align*} f_x(x,y)&=\frac{\partial}{\partial x}(2x^{3}y^{3}+6e^{xy})\ &=2y^{3}\cdot3x^{2}+6e^{xy}\cdot y\ &=6x^{2}y^{3}+6ye^{xy} \end{align*} ]

Step2: Find (f_{xy}(x,y))

Differentiate (f_x(x,y)=6x^{2}y^{3}+6ye^{xy}) with respect to (y). For the first - term (6x^{2}y^{3}), use the power rule ((y^n)^\prime=ny^{n - 1}). For the second - term (6ye^{xy}), use the product rule ((uv)^\prime = u^\prime v+uv^\prime) (where (u = y), (v = e^{xy})). [ \begin{align*} f_{xy}(x,y)&=\frac{\partial}{\partial y}(6x^{2}y^{3}+6ye^{xy})\ &=6x^{2}\cdot3y^{2}+6\left(e^{xy}+y\cdot xe^{xy}\right)\ &=18x^{2}y^{2}+6e^{xy}+6xye^{xy} \end{align*} ]

Step3: Find (f_y(x,y))

Differentiate (f(x,y)=2x^{3}y^{3}+6e^{xy}) with respect to (y). [ \begin{align*} f_y(x,y)&=\frac{\partial}{\partial y}(2x^{3}y^{3}+6e^{xy})\ &=2x^{3}\cdot3y^{2}+6e^{xy}\cdot x\ &=6x^{3}y^{2}+6xe^{xy} \end{align*} ]

Step4: Find (f_{yy}(x,y))

Differentiate (f_y(x,y)=6x^{3}y^{2}+6xe^{xy}) with respect to (y). For the first - term (6x^{3}y^{2}), use the power rule. For the second - term (6xe^{xy}), use the chain rule. [ \begin{align*} f_{yy}(x,y)&=\frac{\partial}{\partial y}(6x^{3}y^{2}+6xe^{xy})\ &=6x^{3}\cdot2y+6x\cdot xe^{xy}\ &=12x^{3}y + 6x^{2}e^{xy} \end{align*} ]

Answer:

(b) (18x^{2}y^{2}+6e^{xy}+6xye^{xy}) (c) (12x^{3}y + 6x^{2}e^{xy})