If(x²+y²)=xy,then dy/dx?
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CORRECT QUESTION:
ANSWER:
[tex] \sf The \ value \ of \ \dfrac{dy}{dx} = \dfrac{ y-4x^{3}-3 y^{2}x}{4x^2y+4y^3-x}[/tex]
GIVEN:
TO FIND:
EXPLANATION:
Let A = (x² + y²)² and B = xy
Take A:
A = (x² + y²)²
Differentiate w.r.t x
[tex] \sf \dfrac{d A }{dx}= \dfrac{d}{dx} (x^{2} + y^{2})^2 [/tex]
[tex]\boxed{ \bold{ \large{\gray{ \dfrac{d}{dx} x^{n} = n {x}^{n - 1}}}}}[/tex]
[tex] \sf \dfrac{d A }{dx}= 2(x^{2} + y^{2})\dfrac{d}{dx}(x^2 + y^2) [/tex]
[tex] \sf \dfrac{d A }{dx}= 2(x^{2} + y^{2})\left(2x+ 2y\dfrac{dy }{dx}\right)[/tex]
[tex] \sf \dfrac{d A }{dx}= (2x^{2} + 2 y^{2})\left(2x+ 2y\dfrac{dy }{dx}\right)[/tex]
[tex] \sf \dfrac{d A }{dx}= 4x^{3} + 4x^2y\left(\dfrac{dy }{dx}\right)+4y^{2}x+4y^3\left(\dfrac{dy }{dx}\right)[/tex]
Take B:
B = xy
Differentiate w.r.t x
[tex] \sf \dfrac{dB}{dx}= \dfrac{d}{dx} xy[/tex]
[tex] \boxed{ \bold {\large{ \gray{\dfrac{d}{dx} uv = uv'+vu'}}}}[/tex]
[tex] \sf \dfrac{dB}{dx}= y \left( \dfrac{d}{dx}x \right) +x \left(\dfrac{d}{dx} y\right)[/tex]
[tex] \boxed{ \bold {\large{ \gray{\dfrac{d}{dx}x = 1}}}}[/tex]
[tex] \sf \dfrac{dB}{dx}= y +x \left(\dfrac{dy}{dx} \right)[/tex]
[tex] \sf We\ know\ that\ A=B[/tex]
Differentiate w.r.t x
[tex] \sf \dfrac{dA}{dx}= \dfrac{dB}{dx}[/tex]
Equate the respective values.
[tex] \sf 4x^{3} + 4x^2y\dfrac{dy }{dx}+4 y^{2}x+4y^3\dfrac{dy }{dx}= y +x\left(\dfrac{dy}{dx} \right)[/tex]
[tex] \sf 4x^2y\dfrac{dy }{dx}+4y^3\dfrac{dy }{dx}-x\left(\dfrac{dy}{dx} \right)= y-4x^{3}-4 y^{2}x [/tex]
Take dy/dx as common.
[tex] \sf \dfrac{dy }{dx}(4x^2y+4y^3-x)= y-4x^{3}-4 y^{2}x [/tex]
[tex] \sf \dfrac{dy}{dx} = \dfrac{ y-4x^{3}-4 y^{2}x}{4x^2y+4y^3-x}[/tex]
[tex] \sf The \ value \ of \ \dfrac{dy}{dx} = \dfrac{ y-4x^{3}-4 y^{2}x}{4x^2y+4y^3-x}[/tex]