Question

Compute fxyx in given multivalued function. f(x, y) = 3x²y + 2xy³ - 4e^(x-y)​

Answer 1

Solution:

Given That:

$\tt \longrightarrow f(x,y) = 3{x}^{2}y +2x {y}^{3} -4 {e}^{x - y}$

Partial differentiating with respect to x, we get:

$\tt \longrightarrow \dfrac{ \partial f}{ \partial x} = \dfrac{ \partial}{ \partial x}(3 {x}^{2}y + 2x {y}^{3} - 4 {e}^{x - y} )$

$\tt \longrightarrow \dfrac{ \partial f}{ \partial x} = \dfrac{ \partial}{ \partial x}(3 {x}^{2}y) + \dfrac{ \partial}{ \partial x} (2x {y}^{3}) - \dfrac{ \partial}{ \partial x} ( 4 {e}^{x - y} )$

$\tt \longrightarrow \dfrac{ \partial f}{ \partial x} = 3y\dfrac{ \partial}{ \partial x}({x}^{2}) +2 {y}^{3} \dfrac{ \partial}{ \partial x} (x) - 4\dfrac{ \partial}{ \partial x} ( {e}^{x - y} )$

$\tt \longrightarrow \dfrac{ \partial f}{ \partial x} = 3y \cdot 2x+2 {y}^{3} \cdot1 - 4 {e}^{x - y} \dfrac{ \partial}{ \partial x} (x -y)$

$\tt \longrightarrow \dfrac{ \partial f}{ \partial x} = 6xy+2 {y}^{3} - 4 {e}^{x - y}$

Partial differentiating with respect to y, we get:

$\tt \longrightarrow \dfrac{ \partial^{2} f}{ \partial x \partial y} = \dfrac{ \partial}{ \partial y}( 6xy+2 {y}^{3} - 4 {e}^{x - y})$

$\tt \longrightarrow \dfrac{ \partial^{2} f}{ \partial x \partial y} = \dfrac{ \partial}{ \partial y}( 6xy)+ \dfrac{ \partial}{ \partial y} (2 {y}^{3}) - \dfrac{ \partial}{ \partial y} (4 {e}^{x - y})$

$\tt \longrightarrow \dfrac{ \partial^{2} f}{ \partial x \partial y} =6x \dfrac{ \partial}{ \partial y}(y)+2 \dfrac{ \partial}{ \partial y} ( {y}^{3}) - 4 {e}^{x - y} \dfrac{ \partial}{ \partial y} (x - y)$

$\tt \longrightarrow \dfrac{ \partial^{2} f}{ \partial x \partial y} =6x+2 \cdot3 {y}^{2} - 4 {e}^{x - y}\times ( - 1)$

$\tt \longrightarrow \dfrac{ \partial^{2} f}{ \partial x \partial y} =6x+6{y}^{2} + 4 {e}^{x - y}$

Partial differentiating with respect to x, we get:

$\tt \longrightarrow \dfrac{ \partial^{3} f}{ \partial x \partial y \partial x} =\dfrac{ \partial}{ \partial x} (6x+6{y}^{2} + 4 {e}^{x - y})$

$\tt \longrightarrow \dfrac{ \partial^{3} f}{ \partial x \partial y \partial x} =6 + 0 + 4 {e}^{x - y}$

$\tt \longrightarrow \dfrac{ \partial^{3} f}{ \partial x \partial y \partial x} =6 + 4 {e}^{x - y}$

Which is our required answer.

Know More:

$\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}$