Question

(x^2+y^2)^2= xy
find dy/dx

Answer 1

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

Given function is

$\rm :\longmapsto\: {( {x}^{2} + {y}^{2} )}^{2} = xy$

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

$\rm :\longmapsto\: \dfrac{d}{dx} {( {x}^{2} + {y}^{2} )}^{2} = \dfrac{d}{dx}xy$

We know

$\boxed{ \bf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

and

$\boxed{ \bf{ \: \dfrac{d}{dx}uv =u\dfrac{d}{dx}v + v\dfrac{d}{dx}u}}$

So, using these two results, we get

$\rm :\longmapsto\:2( {x}^{2} + {y}^{2})\dfrac{d}{dx}( {x}^{2} + {y}^{2}) = x\dfrac{d}{dx}y + y\dfrac{d}{dx}x$

$\rm :\longmapsto\:2( {x}^{2} + {y}^{2})(2x + 2y\dfrac{d}{dx}y) = x\dfrac{dy}{dx} + y \times 1$

$\rm :\longmapsto\:2( {x}^{2} + {y}^{2})(2x + 2y\dfrac{dy}{dx}) = x\dfrac{dy}{dx} + y$

$\rm :\longmapsto\:4x( {x}^{2} + {y}^{2}) + 4y\dfrac{dy}{dx}( {x}^{2} + {y}^{2})= x\dfrac{dy}{dx} + y$

$\rm :\longmapsto\:4y\dfrac{dy}{dx}( {x}^{2} + {y}^{2}) - x\dfrac{dy}{dx} = y - 4x( {x}^{2} + {y}^{2})$

$\rm :\longmapsto\: \bigg(4y( {x}^{2} + {y}^{2}) - x \bigg)\dfrac{dy}{dx} = y - 4x( {x}^{2} + {y}^{2})$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{y - {4x}^{3} - 4x{y}^{2} }{ {4yx}^{2} + {4y}^{3} - x}$

Additional Information :-

$\boxed{ \bf{ \: \dfrac{d}{dx}k = 0}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}x = 1}}$

$\boxed{ \bf{ \: \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } }}$

$\boxed{ \bf{ \: \dfrac{d}{dx}logx = \frac{1}{x}}}$

$\boxed{ \bf{ \: \dfrac{d}{dx} {e}^{x} = {e}^{x}}}$

$\boxed{ \bf{ \: \dfrac{d}{dx} {a}^{x} = {a}^{x} \: loga}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}k \: f(x) \: = \: k \: \dfrac{d}{dx}f(x)}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}sinx = cosx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}cosx = - \: sinx}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}tanx = {sec}^{2}x}}$

$\boxed{ \bf{ \: \dfrac{d}{dx}cotx = - \: {cosec}^{2}x}}$

Answer 2

Answer:

$\orange{ \boxed{ \therefore \bf\dfrac{dy}{dx} = \dfrac{y - {4x}^{3} - 4x{y}^{2} }{ {4yx}^{2} + {4y}^{3} - x}}}$ [/tex]

Explanation :

Given,

$\bf \: {( {x}^{2} + {y}^{2} )}^{2} = xy$

It is a implicit function.

We have to differentiate it w.r.t x

But at first we have to know some basic formula.

$\red{\bf \odot\:{ \bf{ \: \dfrac{d \: {x}^{n} }{dx} = {nx}^{n - 1}}} }\\\\\bf \pink{\odot\:{ \bf{ \: \dfrac{d \: uv}{dx}=u\dfrac{d}{dx}v + v\dfrac{d}{dx}u}}} \\ \\ \green{ \bf \odot\: \frac{d \: (constant)}{dx} = 0} \\ \\ \huge \sf \: now \\ \\ \bf \frac{d \: {( {x}^{2} + {y}^{2} })^{2} }{dx} = \frac{d \: xy}{dx} \\ \\ \bf \implies \: 2( {x}^{2} + {y}^{2})\dfrac{d}{dx}( {x}^{2} + {y}^{2}) = x\dfrac{d}{dx}y + y\dfrac{d}{dx}x \\ \\ </p><p>\bf \implies \:2( {x}^{2} + {y}^{2})(2x + 2y\dfrac{d}{dx}y) = x\dfrac{dy}{dx} + y \times 1 \\ \\ </p><p> \bf \implies \:2( {x}^{2} + {y}^{2})(2x + 2y\dfrac{dy}{dx}) = x\dfrac{dy}{dx} + y \\ \\ </p><p> \bf \implies \:4x( {x}^{2} + {y}^{2}) + 4y\dfrac{dy}{dx}( {x}^{2} + {y}^{2})= x\dfrac{dy}{dx} + y \\ \\ </p><p>\bf \implies \:4y\dfrac{dy}{dx}( {x}^{2} + {y}^{2}) - x\dfrac{dy}{dx} = y - 4x( {x}^{2} + {y}^{2}) \\ \\ \bf \implies \: \{4y( {x}^{2} + {y}^{2}) - x \}\dfrac{dy}{dx} = y - 4x( {x}^{2} + {y}^{2}) \\ \\ \bf \implies \:(4y {x}^{2} + 4 {y}^{3} - x) \frac{dy}{dx} = y - 4 {x}^{3} + 4x {y}^{2} \\ \\ \bf \implies \: - (x - 4 {y}^{3} - 4y {x}^{2} ) \frac{dy}{dx} = - (4 {x}^{3} - 4x {y}^{2} - y) \\ \\ \bf\implies \:\dfrac{dy}{dx} = \dfrac{y - {4x}^{3} - 4x{y}^{2} }{ {4yx}^{2} + {4y}^{3} - x} \\ \\ \\ \orange{ \boxed{ \therefore \bf\dfrac{dy}{dx} = \dfrac{y - {4x}^{3} - 4x{y}^{2} }{ {4yx}^{2} + {4y}^{3} - x}}}$ [/tex]