Question

Calcule a derivadas parciais da segunda ordem da função. f(x,y)=sen(3x+y²)

Answer 1

Regra da cadeia
1º em relação a x
f(x) = cos(3x + y²)*3 = 3cos(3x + y²)
= -3sen(3x + y²)*3 = - 9 sen(3x + y²)
2º em relação a y
f(y) = cos(3x + y²)*2y = 2y cos(3x + y2)
regra do produto
2cos(3x + y²) + 2y[-sen(3x + y²)*2y]
= 2cos(3x + y²) - 4y²sen(3x + y²)
= 2[cos(3x + y²) - 2y²sen(3x + y²)]

f(x,y) = f(y,x)
f(x) = 3cos(3x + y²)
f(x,y) = -3sen(3x + y²)*2y = -6ysen(3x + y²)
derivadas de 2ª ordem
f(x,x) = -9sen(3x + y²)
f(y,y) = 2[cos(3x = y²) -2y²sen(3x + y²)]
f(x,y) = f(y,x) = -6sen(3x + y²)

Answer 2

✅ Após resolver os cálculos, concluímos que as derivadas parciais de segundo ordem da referida função são, respectivamente:

                  $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf f_{xx}(x, y) = -9\sin(3x + y^{2})\:\:\:}}\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf f_{yy}(x, y) = 2\left[\cos(3x + y^{2} ) - 2y^{2}\sin(3x + y^{2})\right]\:\:\:}}\end{gathered} }$

Seja a função em duas variáveis:

              $\Large\displaystyle\begin{gathered} f(x, y) = \sin(3x + y^{2})\end{gathered}$

Para calcular as derivadas parciais de segunda ordem da função devemos:

• Calcular a derivada parcial de segunda ordem em termos de "x".

        $\Large\displaystyle\begin{gathered} f_{xx}(x, y) = f_{x}\left[f_{x}(x, y)\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = f_{x}\left[\cos(3x + y^{2})\cdot1\cdot3\cdot x^{1 - 1}\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = f_{x}\left[\cos(3x + y^{2})\cdot3\cdot x^{0}\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = f_{x}\left[\cos(3x + y^{2})\cdot3\cdot 1\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = f_{x}\left[\cos(3x + y^{2})\cdot3\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = f_{x}\left[3\cos(3x + y^{2})\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = 3\cdot\left[-\sin(3x + y^{2})\cdot1\cdot3\cdot x^{1 - 1}\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = 3\cdot\left[-\sin(3x + y^{2})\cdot3\cdot x^ {0}\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = 3\cdot\left[-\sin(3x + y^{2})\cdot3\cdot 1\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = 3\cdot\left[-\sin(3x + y^{2})\cdot3\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = 3\cdot\left[-3\sin(3x + y^{2})\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = -9\sin(3x + y^{2})\end{gathered}$

• Calcular a derivada parcial de segunda ordem em termos de "y".

        $\Large\displaystyle\begin{gathered} f_{yy}(x, y) = f_{y}\left[f_{y}(x, y)\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = f_{y}\left[\cos(3x + y^{2})\cdot2\cdot 1\cdot y^{2 - 1}\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = f_{y}\left[\cos(3x + y^{2})\cdot2\cdot y\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = f_{y}\left[2y\cos(3x + y^{2})\right]\end{gathered}$

                           $\large\displaystyle\begin{gathered} = 2\cdot\cos(3x + y^{2}) + 2y\cdot\left[-\sin(3x + y^{2})\cdot2y\right]\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = 2\cos(3x + y^{2}) - 4y^{2}\sin(3x + y^{2})\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = 2\left[\cos(3x + y^{2} ) - 2y^{2}\sin(3x + y^{2})\right]\end{gathered}$

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

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$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$