Question

$\rm If \: {x}^{3} + {y}^3 - {x}^3 {y}^{3} z = 0 , \: find \\ \rm \frac{ \partial z}{\partial x} \: and \: \frac{\partial z}{\partial y } \: when \: x = y = 1$​

Answer 1

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

Given function is

$\rm \: {x}^{3} + {y}^3 - {x}^3 {y}^{3} z = 0$

can be rewritten as

$\rm \: {x}^{3} + {y}^3 = {x}^3 {y}^{3} z$

$\rm \: z = \dfrac{ {x}^{3} + {y}^{3} }{ {x}^{3} {y}^{3} }$

$\rm \: z = \dfrac{ {x}^{3}}{ {x}^{3} {y}^{3} } + \dfrac{ {y}^{3} }{ {x}^{3} {y}^{3} }$

$\rm \: z = \dfrac{1}{ {y}^{3} } + \dfrac{1}{ {x}^{3} }$

can be further rewritten as

$\rm \: z = {y}^{ - 3} + {x}^{ - 3}$

On differentiating partially w. r. t. x, we get

$\rm \: \dfrac{\partial }{\partial x}z \: = \: \dfrac{\partial }{\partial x}( {y}^{ - 3} + {x}^{ - 3})$

$\rm \: \dfrac{\partial z}{\partial x} = 0 + ( - 3) {x}^{ - 3 - 1}$

$\rm\implies \:\boxed{ \rm{ \:\dfrac{\partial z}{\partial x} = - 3 {x}^{ - 4} \: }}$

So,

$\bf\implies \:\dfrac{\partial z}{\partial x}_{(x=1,y=1)} = - 3 {( - 1)}^{ - 4} = - 3$

Now, Consider again

$\rm \: z = {y}^{ - 3} + {x}^{ - 3}$

On differentiating partially w. r. t. y, we get

$\rm \: \dfrac{\partial }{\partial y}z \: = \: \dfrac{\partial }{\partial y}( {y}^{ - 3} + {x}^{ - 3})$

$\rm \: \dfrac{\partial z}{\partial y} = ( - 3) {y}^{ - 3 - 1} + 0$

$\rm\implies \:\boxed{ \rm{ \:\dfrac{\partial z}{\partial y} = - 3 {y}^{ - 4} \: }}$

So,

$\bf\implies \:\dfrac{\partial z}{\partial y}_{(x=1,y=1)} = - 3 {( - 1)}^{ - 4} = - 3$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Answer:

Solution

Given function is

x ^ 2 + y ^ 2 - x ^ 3 * y ^ 3 * x = 0

cen be rewritten as

x ^ 2 + y ^ 2 - x ^ 3 * y ^ 2 * y

z = (x ^ 3 + y ^ 3)/(x ^ 3 * y ^ 3)

z - (x ^ 2)/(x ^ 3 * y ^ 3) + (y ^ 3)/(pi ^ 2 * y ^ 3)

x = 1/(y ^ 3) + 1/(x ^ 3)

can be further rewritten as

x = y ^ - 3 + x ^ - 3 On differentiating partialy w.rt. x, we

got partial partial x x= partial partial x (y^ -3 +x^ -4 ).

partial x partial x -0+(-3)x^ -3-1

Rightarrow boxed partial z partial z =-3x^ 4

- partial z partial x (y-1,y-1) --3(-1)^ -4 --3

Now, Consider again

x = y ^ - 2 + x ^ - 3 On differentiating partialy w. r. t. y. we get

partial partial y y- partial partial y (y^ -3 +x^ -3 )

partial x partial y =(-3)x^ -3-1 +0

boxed partial x partial y =-3y^ -4

partial z partial y :=1,y-1) =-3(-1)^ -4 =-3