Respuesta :
Answer:
Given That:
[tex]{ \large\longrightarrow { \rm{z = {x}^{2} + {y}^{2} }}}[/tex]
Partial differentiating both sides with respect to x, we get:
[tex]{ \large\longrightarrow { \rm{ \frac{∂z}{∂x} = 2x}}}[/tex]
Partial differentiating again with respect to x, we get:
[tex]{ \large\longrightarrow {{ \rm{ \frac{ {∂}^{2}z }{∂ {x}^{2} } \: = 2 \: \: \: \: \quad -(i) }}}}[/tex]
Consider again:
[tex]{ \large\longrightarrow {{ \rm{ z = {x}^{2} + {y}^{2} }}}}[/tex]
Partial differentiating both sides with respect to y, we get:
[tex]{ \large\longrightarrow { { \rm{ \frac{∂z}{∂y} = 2y}}}}[/tex]
Now, partial differentiating both sides with respect to x, we get:
[tex]{ \large\longrightarrow{ \rm{ \frac{ {∂}^{2}z }{∂x \:∂y } = 0 \: \: \: \quad - (ii) }}}[/tex]
Consider again:
[tex]{ \large\longrightarrow { \rm{ \frac{∂z}{∂y} = 2y}}}[/tex]
Partial differentiating both sides with respect to y, we get:
[tex]{ \large\longrightarrow{ \rm{ \frac{ {∂}^{2}z }{∂ {y}^{2} } = 2 \: \: \: \: \: \quad - (iii) }}}[/tex]
Now, consider Left Hand Side, we get:
[tex]{ \large{ \rm{ \longrightarrow {x}^{2} \frac{ {∂}^{2}z }{∂ {x}^{2} } + 2xy \frac{ {∂}^{2} x}{∂x \:∂y } + {y}^{2} \frac{ {∂}^{2}z }{∂ {y}^{2} } }}}[/tex]
From Equation (i),(ii) and (iii), we can write:
[tex]\longrightarrow{ \large{ \rm{ {2x}^{2} + 0 + {2y}^{2} }}}[/tex]
[tex]\longrightarrow \: { \large{ \rm{2( {x}^{2} + {y}^{2} ) }}}[/tex]
[tex]\longrightarrow{ \large{ \rm{2z}}}[/tex]
Therefore,
[tex]{ \large{\longrightarrow \: { \rm{{x}^{2} \frac{ {∂}^{2} z}{∂ {x}^{2} } + 2xy \frac{ {∂}^{2}x }{∂x \:∂y } + {y}^{2} \frac{ {∂}^{2}z }{∂ {y}^{2} } = 2z }}}}[/tex]
Hence Proved!!
[tex] \: [/tex]
Learn More:-
[tex]\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}}[/tex]
Refer to the attachments for answer
Used Concept :-
- [tex]{\boxed{\bf{\dfrac{d}{dx}(x^n)=nx^{n-1}}}}[/tex]
