partial derviatives: f(x, y) = sin(3xy)

[refid]

Hi

Not sure if im doing this right:

\(\displaystyle f (x,y) = sin(3xy); Find {{\partial ^3 f} \over {\partial y^2 \partial x}}\).

my work:

\(\displaystyle {{\partial f} \over {\partial y}} = cos(3xy) 3x\)


\(\displaystyle {{\partial ^2 f} \over {\partial y^2}}= -sin(3xy) 9x^2\)


\(\displaystyle {{\partial ^3 f} \over {\partial y^2 \partial x}} = -18x(-sin(3xy)) - 9x^2(cos(3xy) 3x\)

 

[pka]

In doing higher partial derivatives, one must be very careful about order. Now in your problem the expression is so ‘well behaved’ we do not need to worry.
However, you should learn the correct order.
\(\displaystyle \L
\frac{{\partial ^3 f}}{{\partial ^2 y\partial x}} = \frac{\partial }{{\partial y}}\left( {\frac{\partial }{{\partial y}}} \right)\left[ {\frac{{\partial f}}{{\partial x}}} \right]\)

Please note that the partial with respect to x is done first.
If you are familiar with the more modern ‘subscript notation’ then
\(\displaystyle \L
\frac{{\partial ^3 f}}{{\partial ^2 y\partial x}} = f_{xyy}\)

 

[refid]

ahh i got my notation mixed up

 

[refid]

When is statment true?

\(\displaystyle \L
\frac{{\partial ^3 f}}{{\partial ^2 y\partial x}} =

\frac{{\partial ^3 f}}{{\partial ^2 x\partial y}}\)

All I know is that if the funtion is continous.. so are the partial derviative...

Would i prove the orginial function:

\(\displaystyle \L f( x , y ).... defined ....at ....... x(0) y(0)\)

\(\displaystyle \L\\\lim_{(x,y)\to\ x(0),y(0)} f( x , y ) ....... Exists\)

\(\displaystyle \L\\\lim_{(x,y)\to\ x(0),y(0)} f( x , y ) = f( x(0) , y(0) )\)

is it suffice to say that on an exam?

 

[pka]

In general we require all orders of the partial derivatives must exist and be continuous.