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this is similar to section 6.1 problem 34: for ( f(x,y)=2x^{3}y^{3}+6e^…

Question

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

Explanation:

Step1: Find the first - order partial derivative \(f_x(x,y)\)

Use the power rule \((x^n)^\prime=nx^{n - 1}\) and the chain rule \((e^{u})^\prime=e^{u}\cdot u^\prime\) (where \(u = xy\)).
For the term \(2x^{3}y^{3}\), the derivative with respect to \(x\) is \(2y^{3}\cdot3x^{2}=6x^{2}y^{3}\).
For the term \(6e^{xy}\), using the chain rule, the derivative with respect to \(x\) is \(6e^{xy}\cdot y = 6ye^{xy}\).
So, \(f_x(x,y)=6x^{2}y^{3}+6ye^{xy}\).

Step2: Find the second - order partial derivative \(f_{xx}(x,y)\)

Differentiate \(f_x(x,y)\) with respect to \(x\) again.
For the term \(6x^{2}y^{3}\), using the power rule \((ax^{n})^\prime=anx^{n - 1}\) (here \(a = 6y^{3}\), \(n = 2\)), the derivative is \(6y^{3}\cdot2x=12xy^{3}\).
For the term \(6ye^{xy}\), using the chain rule (let \(u = xy\), then \((6ye^{u})^\prime=6y\cdot e^{u}\cdot y\)). So the derivative is \(6y^{2}e^{xy}\).

Answer:

\(f_{xx}(x,y)=12xy^{3}+6y^{2}e^{xy}\)