Question

If(x²+y²)=xy,then dy/dx?

Answer 1

CORRECT QUESTION:

• If (x² + y²)² = xy, then dy/dx.

ANSWER:

$\sf The \ value \ of \ \dfrac{dy}{dx} = \dfrac{ y-4x^{3}-3 y^{2}x}{4x^2y+4y^3-x}$

GIVEN:

• (x² + y²)² = xy.

TO FIND:

• The value of dy/dx.

EXPLANATION:

Let A = (x² + y²)² and B = xy

Take A:

A = (x² + y²)²

Differentiate w.r.t x

$\sf \dfrac{d A }{dx}= \dfrac{d}{dx} (x^{2} + y^{2})^2$

$\boxed{ \bold{ \large{\gray{ \dfrac{d}{dx} x^{n} = n {x}^{n - 1}}}}}$

$\sf \dfrac{d A }{dx}= 2(x^{2} + y^{2})\dfrac{d}{dx}(x^2 + y^2)$

$\sf \dfrac{d A }{dx}= 2(x^{2} + y^{2})\left(2x+ 2y\dfrac{dy }{dx}\right)$

$\sf \dfrac{d A }{dx}= (2x^{2} + 2 y^{2})\left(2x+ 2y\dfrac{dy }{dx}\right)$

$\sf \dfrac{d A }{dx}= 4x^{3} + 4x^2y\left(\dfrac{dy }{dx}\right)+4y^{2}x+4y^3\left(\dfrac{dy }{dx}\right)$

Take B:

B = xy

Differentiate w.r.t x

$\sf \dfrac{dB}{dx}= \dfrac{d}{dx} xy$

$\boxed{ \bold {\large{ \gray{\dfrac{d}{dx} uv = uv'+vu'}}}}$

$\sf \dfrac{dB}{dx}= y \left( \dfrac{d}{dx}x \right) +x \left(\dfrac{d}{dx} y\right)$

$\boxed{ \bold {\large{ \gray{\dfrac{d}{dx}x = 1}}}}$

$\sf \dfrac{dB}{dx}= y +x \left(\dfrac{dy}{dx} \right)$

$\sf We\ know\ that\ A=B$

Differentiate w.r.t x

$\sf \dfrac{dA}{dx}= \dfrac{dB}{dx}$

Equate the respective values.

$\sf 4x^{3} + 4x^2y\dfrac{dy }{dx}+4 y^{2}x+4y^3\dfrac{dy }{dx}= y +x\left(\dfrac{dy}{dx} \right)$

$\sf 4x^2y\dfrac{dy }{dx}+4y^3\dfrac{dy }{dx}-x\left(\dfrac{dy}{dx} \right)= y-4x^{3}-4 y^{2}x$

Take dy/dx as common.

$\sf \dfrac{dy }{dx}(4x^2y+4y^3-x)= y-4x^{3}-4 y^{2}x$

$\sf \dfrac{dy}{dx} = \dfrac{ y-4x^{3}-4 y^{2}x}{4x^2y+4y^3-x}$

$\sf The \ value \ of \ \dfrac{dy}{dx} = \dfrac{ y-4x^{3}-4 y^{2}x}{4x^2y+4y^3-x}$