Question 205262
Having trouble can not figure this out 
for f(x,y)=3x^2y^5+6x^6y^2 
compute 
{{{f[xx]}}}(1,-1) = 
{{{f[xy]}}}(2,2) = 
{{{f[yy]}}}(-1,-1) = 
thanks so much for help
<pre><font size = 4 color = "indigo"><b>

Find the first partials:

{{{f(x,y)=3x^2y^5+6x^6y^2}}}

To find {{{f[x]}}}(x,y), consider y to be
a constant, in the first term {{{3y^5}}} is considered
constant, and in the second term {{{6y^2}}} is considered
constant.  If you like, you can rewrite {{{f(x,y)}}} so what 
is to be held constant is in parentheses

{{{f(x,y)=(3y^5)x^2+(6y^2)x^6}}}

Then we use the ordinary derivative formulas,
considering what is in parentheses as constant:

{{{f[x]}}}{{{"(x,y)"}}}={{{ 2(3y^5)x+6(6y^2)x^5 }}}

simplifying,

{{{f[x]}}}{{{"(x,y)"}}}={{{ 6y^5x+36y^2x^5 }}}

get it in alphabetical order

{{{f[x]}}}{{{"(x,y)"}}}={{{6xy^5+36x^5y^2}}}

----------------

{{{f(x,y)=3x^2y^5+6x^6y^2}}}

To find {{{f[y]}}}(x,y), consider x to be
a constant, in the first term {{{3x^2}}} is considered
constant, and in the second term {{{6x^6}}} is considered
constant.  If you like, you can rewrite {{{f(x,y)}}} so what 
is to be held constant is in parentheses

{{{f(x,y)=(3x^2)y^5+(6x^6)y^2}}}

Then we use the ordinary derivative formulas,
considering what is in parentheses as constant:

{{{f[y]}}}{{{"(x,y)"}}}={{{ 5(3x^2)y^4+2(6x^6)y }}}

simplifying,

{{{f[y]}}}{{{"(x,y)"}}}={{{15x^2y^4+12x^6y }}}

===============================================

Find the second partial derivative {{{f[xx]}}}, which
is the partial derivative with respect to x of 
the partial derivative with respect to x.

Start with the partial derivative with respect to x.

{{{f[x]}}}{{{"(x,y)"}}}={{{6xy^5+36x^5y^2}}}

To find {{{f[xx]}}}(x,y), consider y to be
a constant, in the first term {{{6y^5}}} is considered
constant, and in the second term {{{36y^2}}} is considered
constant.  If you like, you can rewrite {{{f[x]}}} so what 
is to be held constant is in parentheses

{{{f[x]}}}{{{"(x,y)"}}}={{{(6y^5)x+(36y^2)x^5}}}

Then we use the ordinary derivative formulas,
considering what is in parentheses as constant:

{{{f[xx]}}}{{{"(x,y)"}}}={{{ (6y^5)+5(36y^2)x^4 }}}

simplifying,

{{{f[xx]}}}{{{"(x,y)"}}}={{{ 6y^5+180y^2x^4 }}}

get it in alphabetical order

{{{f[xx]}}}{{{"(x,y)"}}}={{{6y^5+180x^4y^2}}}

And substituting

{{{f[xx]}}}{{{"(1,-1)"}}}={{{6(-1)^5+180(1)^4(-1)^2}}}

{{{f[xx]}}}{{{"(1,-1)"}}}={{{6(-1)+180(1)(1)}}}

{{{f[xx]}}}{{{"(1,-1)"}}}={{{-6+180}}}

{{{f[xx]}}}{{{"(1,-1)"}}}={{{174}}}

=============================================== 

===============================================

Find the second partial derivative {{{f[xy]}}}, which
is the partial derivative with respect to y of 
the partial derivative with respect to x.

Start with the partial derivative with respect to x.

{{{f[x]}}}{{{"(x,y)"}}}={{{6xy^5+36x^5y^2}}}

To find {{{f[xy]}}}(x,y), consider x to be
a constant, in the first term {{{6x}}} is considered
constant, and in the second term {{{36x^5}}} is considered
constant.  If you like, you can rewrite {{{f[x]}}} so what 
is to be held constant is in parentheses

{{{f[x]}}}{{{"(x,y)"}}}={{{(6x)y^5+(36x^5)y^2}}}

Then we use the ordinary derivative formulas,
considering what is in parentheses as constant:

{{{f[xy]}}}{{{"(x,y)"}}}={{{5(6x)y^4+2(36x^5)y }}}

simplifying,

{{{f[xy]}}}{{{"(x,y)"}}}={{{30xy^4+72x^5y }}}

Substituting:

{{{f[xy]}}}{{{"(2,2)"}}}={{{30(2)(2)^4+72(2)^5(2) }}}

{{{f[xy]}}}{{{"(2,2)"}}}={{{30(2)(16)+72(32)(2) }}}

{{{f[xy]}}}{{{"(2,2)"}}}={{{960+4608 }}}

{{{f[xy]}}}{{{"(2,2)"}}}={{{5568}}}

=============================================== 

{{{f[y]}}}{{{"(x,y)"}}}={{{15x^2y^4+12x^6y }}}

Find the second partial derivative {{{f[yy]}}}, which
is the partial derivative with respect to y of 
the partial derivative with respect to y.

Start with the partial derivative with respect to y.

{{{f[y]}}}{{{"(x,y)"}}}={{{15x^2y^4+12x^6y}}}

To find {{{f[yy]}}}(x,y), consider x to be
a constant, in the first term {{{15x^2}}} is considered
constant, and in the second term {{{12x^6}}} is considered
constant.  If you like, you can rewrite {{{f[y]}}} so what 
is to be held constant is in parentheses

{{{f[y]}}}{{{"(x,y)"}}}={{{(15x^2)y^4+(12x^6)y}}}

Then we use the ordinary derivative formulas,
considering what is in parentheses as constant:

{{{f[yy]}}}{{{"(x,y)"}}}={{{4(15x^2)y^3+(12x^6) }}}

simplifying,

{{{f[yy]}}}{{{"(x,y)"}}}={{{50x^2y^3+12x^6 }}}

Substituting:

{{{f[yy]}}}{{{"(-1,-1)"}}}={{{50(-1)^2(-1)^3+12(-1)^6 }}}

{{{f[yy]}}}{{{"(2,2)"}}}={{{50(1)(-1)+12(1) }}}

{{{f[yy]}}}{{{"(2,2)"}}}={{{-50+12 }}}

{{{f[yy]}}}{{{"(2,2)"}}}={{{-38}}}

Edwin</pre>