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Section 13.4 : Higher Order Partial Derivatives

3. Find all 2nd order derivatives for the following function.

\[g\left( {u,v} \right) = {u^3}{v^4} - 2u\sqrt {{v^3}} + {u^6} - \sin \left( {3v} \right)\]

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First, we know we’ll need the two 1st order partial derivatives. Here they are,

\[{g_u} = 3{u^2}{v^4} - 2{v^{\frac{3}{2}}} + 6{u^5}\hspace{0.5in}{g_v} = 4{u^3}{v^3} - 3u{v^{\frac{1}{2}}} - 3\cos \left( {3v} \right)\] Show Step 2

Now let’s compute each of the second order partial derivatives.

\[\begin{align*}{g_{u\,u}} & = {\left( {{g_u}} \right)_u} = 6u{v^4} + 30{u^4}\\ {g_{u\,v}} & = {\left( {{g_u}} \right)_v} = 12{u^2}{v^3} - 3{v^{\frac{1}{2}}}\\ {g_{v\,u}} & = {g_{u\,v}} = 12{u^2}{v^3} - 3{v^{\frac{1}{2}}}\hspace{0.5in}{\mbox{by Clairaut's Theorem}}\\ {g_{v\,v}} & = {\left( {{g_v}} \right)_v} = 12{u^3}{v^2} - \frac{3}{2}u{v^{ - \,\,\frac{1}{2}}} + 9\sin \left( {3v} \right)\end{align*}\]