Question

if u =x+y/x-y, find del u /delx, del u/del y in topper answer​

Answer 1

Question :-

$\rm \:If \: u \: = \: \dfrac{x + y}{x - y}, \: find \: \dfrac{\partial u}{\partial x} \: and \: \dfrac{\partial u}{\partial y}$

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

Given expression is

$\rm \: u \: = \: \dfrac{x + y}{x - y}$

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

$\rm \: \dfrac{\partial }{\partial x} u \: = \: \dfrac{\partial }{\partial x} \: \dfrac{x + y}{x - y}$

We know,

$\boxed{ \rm{ \:\dfrac{d}{dx} \frac{u}{v} \: = \: \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } \: \: }}$

So, using this, we get

$\rm \: \dfrac{\partial u}{\partial x} = \dfrac{(x - y)\dfrac{\partial }{\partial x} (x + y) - (x + y)\dfrac{\partial }{\partial x}(x - y)}{ {(x - y)}^{2} }$

$\rm \: \dfrac{\partial u}{\partial x} = \dfrac{(x - y)(1 + 0) - (x + y)(1 - 0)}{ {(x - y)}^{2} }$

$\rm \: \dfrac{\partial u}{\partial x} = \dfrac{(x - y) - (x + y)}{ {(x - y)}^{2} }$

$\rm \: \dfrac{\partial u}{\partial x} = \dfrac{x - y - x - y}{ {(x - y)}^{2} }$

$\rm \: \dfrac{\partial u}{\partial x} = \dfrac{ - 2y}{ {(x - y)}^{2} }$

Hence,

$\rm\implies \: \boxed{ \rm{ \:\: \rm \: \dfrac{\partial u}{\partial x} = \dfrac{ - 2y}{ {(x - y)}^{2} } \: \: }}$

Again Consider

$\rm \: u \: = \: \dfrac{x + y}{x - y}$

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

$\rm \: \dfrac{\partial }{\partial y} u \: = \: \dfrac{\partial }{\partial y} \: \dfrac{x + y}{x - y}$

$\rm \: \dfrac{\partial u}{\partial y} = \dfrac{(x - y)\dfrac{\partial }{\partial y} (x + y) - (x + y)\dfrac{\partial }{\partial y}(x - y)}{ {(x - y)}^{2} }$

$\rm \: \dfrac{\partial u}{\partial y} = \dfrac{(x - y)(0 + 1) - (x + y)(0 - 1)}{ {(x - y)}^{2} }$

$\rm \: \dfrac{\partial u}{\partial y} = \dfrac{(x - y) + (x + y)}{ {(x - y)}^{2} }$

$\rm \: \dfrac{\partial u}{\partial y} = \dfrac{2x}{ {(x - y)}^{2} }$

Hence,

$\rm\implies \:\boxed{ \rm{ \: \: \rm \: \dfrac{\partial u}{\partial y} = \dfrac{2x}{ {(x - y)}^{2} } \: \: }}$

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

Additional Information :-

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