A bit of implicit differentiation never hurt anyone

Calculus Level pending

Use implicit differentiation to find y ′ y'

( 2 x + y ) 4 + 3 x 2 + 3 y 2 = x y + 1 (2x+y)^4+3x^2+3y^2=\frac{x}{y}+1

y ′ = 4 y 2 ( 2 x + y ) 3 + 6 y 3 + x y − 8 y 2 ( 2 x + y ) 3 − 6 x y 2 y'=\frac{4y^2(2x+y)^3+6y^3+x}{y-8y^2(2x+y)^3-6xy^2} y ′ = y − 8 y 2 ( 2 x + y ) 3 − 6 x y 2 4 y 2 ( 2 x + y ) 3 + 6 y 3 + x y'=\frac{y-8y^2(2x+y)^3-6xy^2}{4y^2(2x+y)^3+6y^3+x} y ′ = x ( 2 x + y ) 4 + 3 x 2 + 3 y 2 y'=\frac{x}{(2x+y)^4+3x^2+3y^2} y ′ = y ( ( 2 x + y ) 4 + 3 x 2 + 3 y 2 − 1 ) y'=y((2x+y)^4+3x^2+3y^2-1)

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Tom Gallagher by Tom Gallagher , 22, United Kingdom

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1 solution

Tom Gallagher
Feb 1, 2015

( 2 x + y ) 4 + 3 x 2 + 3 y 2 = x y + 1 (2x+y)^4+3x^2+3y^2=\frac{x}{y}+1

4 ( 2 x + y ) 3 ( 2 + y ′ ) + 6 x + 6 y y ′ = y − x y ′ y 2 4(2x+y)^3(2+y')+6x+6yy'=\frac{y-xy'}{y^2}

8 ( 2 x + y ) 3 + 4 y ′ ( 2 x + y ) 3 + 6 x + 6 y y ′ = 1 y − x y ′ y 2 8(2x+y)^3+4y'(2x+y)^3+6x+6yy'=\frac{1}{y}-\frac{xy'}{y^2}

y ′ ( 4 ( 2 x + y ) 3 + 6 y + x y 2 ) = 1 y − 8 ( 2 x + y ) 3 − 6 x y'(4(2x+y)^3+6y+\frac{x}{y^2})=\frac{1}{y}-8(2x+y)^3-6x

y ′ ( 4 y 2 ( 2 x + y ) 3 + 6 y 3 + x y 2 ) = y − 8 y 2 ( 2 x + y ) 3 − 6 x y 2 y 2 y'(\frac{4y^2(2x+y)^3+6y^3+x}{y^2})=\frac{y-8y^2(2x+y)^3-6xy^2}{y^2}

∴ y ′ = y − 8 y 2 ( 2 x + y ) 3 − 6 x y 2 4 y 2 ( 2 x + y ) 3 + 6 y 3 + x \therefore y'=\frac{y-8y^2(2x+y)^3-6xy^2}{4y^2(2x+y)^3+6y^3+x}

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