Question

$\rm Given \: \red{y=f(x_1,x_2)=3x^2_1+x_1x_2+4x_2^2}, \\ \rm find \: their \: partial \: derivatives.$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: y=f(x_1,x_2)=3x^2_1+x_1x_2+4x_2^2$

On differentiating partially w. r. t. $\rm \: x_1$, we get

$\rm \: \dfrac{\partial }{\partial x_1}y =\dfrac{\partial }{\partial x_1}(3x^2_1+x_1x_2+4x_2^2)$

$\rm \: \dfrac{\partial y}{\partial x_1} = 3\dfrac{\partial }{\partial x_1} {x_1}^{2} + x_2\dfrac{\partial }{\partial x_1}x_1 + \dfrac{\partial }{\partial x_1} {4x_2}^{2}$

$\rm \: \dfrac{\partial y}{\partial x_1} = 3(2x_1) + x_2 \times 1 + 0$

$\rm\implies \:\bf \: \dfrac{\partial y}{\partial x_1} =6x_1+ x_2$

Now, On differentiating partially w. r. t. $\rm \: x_2$, we have

$\rm \: \dfrac{\partial }{\partial x_2}y =\dfrac{\partial }{\partial x_2}(3x^2_1+x_1x_2+4x_2^2)$

$\rm \: \dfrac{\partial y}{\partial x_2} = \dfrac{\partial }{\partial x_2} 3{x_1}^{2} + x_1\dfrac{\partial }{\partial x_2}x_2 +4 \dfrac{\partial }{\partial x_2} {x_2}^{2}$

$\rm \: \dfrac{\partial y}{\partial x_2} = 0 + x_1 \times 1 + 4(2x_2)$

$\rm\implies \:\bf \: \dfrac{\partial y}{\partial x_2} =x_1+ 8x_2$

Now,

$\rm \: \dfrac{\partial ^{2}y}{\partial x_1 ^{2} }$

$\rm \: = \: \dfrac{\partial }{\partial x_1}\bigg(\dfrac{\partial y}{\partial x_1}\bigg)$

$\rm \: = \: \dfrac{\partial }{\partial x_1}(6x_1 + x_2)$

$\rm \: = \: 6\dfrac{\partial }{\partial x_1}x_1 + \dfrac{\partial }{\partial x_1}x_2$

$\rm \: = \: 6 \times 1 + 0$

$\rm \: = \: 6$

$\bf\implies \: \dfrac{\partial ^{2}y}{\partial x_1 ^{2} } = 6$

Now,

$\rm \: \dfrac{\partial ^{2}y}{\partial x_2 ^{2} }$

$\rm \: = \: \dfrac{\partial }{\partial x_2}\bigg(\dfrac{\partial y}{\partial x_2}\bigg)$

$\rm \: = \: \dfrac{\partial }{\partial x_2}(x_1 + 8x_2)$

$\rm \: = \: \dfrac{\partial }{\partial x_2}x_1 + 8\dfrac{\partial }{\partial x_2}x_2$

$\rm \: = \: 0 + 8 \times 1$

$\rm \: = \: 8$

$\bf\implies \: \dfrac{\partial ^{2}y}{\partial x_2 ^{2} } = 8$

Now,

$\rm \: \dfrac{\partial^{2}y }{\partial x_1 \: \partial x_2}$

$\rm \: = \: \dfrac{\partial }{\partial x_1}\bigg(\dfrac{\partial y}{\partial x_2} \bigg)$

$\rm \: = \: \dfrac{\partial }{\partial x_1}(x_1 + 8x_2)$

$\rm \: = \: \dfrac{\partial }{\partial x_1}x_1 + \dfrac{\partial }{\partial x_1}8x_2$

$\rm \: = \: 1 + 0$

$\rm \: = \: 1$

$\bf\implies \: \dfrac{\partial^{2}y }{\partial x_1 \: \partial x_2} = 1$

Answer 2

Solution :-

Given: $\rm y=f\left(x_{1}, x_{2}\right)=3 x_{1}^{2}+x_{1} x_{2}+4 x_{2}^{2}$

To find: Partial derivatives with respect to $\rm x_{1}$ and $\rm x_{2}$

$\[ \begin{aligned} \rm f_{x_{1}}=\frac{\partial y}{\partial x_{1}} & \rm=\frac{\partial}{\partial x_{1}}\left(3 x_{1}^{2}+x_{1} x_{2}+4 x_{2}^{2}\right) \\\\ & \rm=\frac{\partial}{\partial x_{1}} 3 x_{1}^{2}+\frac{\partial}{\partial x_{1}} x_{1} x_{2}+\frac{\partial}{\partial x_{1}} 4 x_{2}^{2} \\ \\ & \rm=3 \frac{\partial}{\partial x_{1}} x_{1}^{2}+x_{2} \frac{\partial}{\partial x_{1}} x_{1}+0 \\ \\ & \rm=3.2 x_{1}+x_{2} .1 \\ \\ \color{green}\rm f_{x_{1}}=\frac{\partial y}{\partial x_{1}} & \color{green}\rm=6 x_{1}+x_{2} \end{aligned} \]$

$\[ \begin{aligned} \rm f_{x_{2}}=\frac{\partial y}{\partial x_{2}} & \rm=\frac{\partial}{\partial x_{2}}\left(3 x_{1}^{2}+x_{1} x_{2}+4 x_{2}^{2}\right) \\ \\ & \rm=\frac{\partial}{\partial x_{2}} 3 x_{1}^{2}+\frac{\partial}{\partial x_{2}} x_{1} x_{2}+\frac{\partial}{\partial x_{2}} 4 x_{2}^{2} \\ \\ & \rm=0+x_{1} \frac{\partial}{\partial x_{2}} x_{2}+4 \frac{\partial}{\partial x_{2}} x_{2}^{2} \\ \\ & \rm=x_{1} .1+4.2 x_{2} \\ \\ \color{blue} \rm f_{x_{2}}=\frac{\partial y}{\partial x_{2}} &\color{blue} \rm=x_{1}+8 x_{2} \end{aligned} \]$