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Finding derivative of Implicit functions
Last updated at April 13, 2021 by Teachoo
Misc 14 If π₯ β(1+π¦)+π¦ β(1+π₯) = 0 , for β1 < π₯ < 1, prove that ππ¦/ππ₯ = (β1)/(1 + π₯)2 π₯ β(1+π¦)+π¦ β(1+π₯) = 0 π₯ β(1+π¦) = β π¦ β(1+π₯) Squaring both sides (π₯β(1+π¦) )^2 = (βπ¦ β(1+π₯))^2 π₯^2 (β(1+π¦ ) )^2 = (βπ¦)^2 (β(1+π₯))^2 π₯^2 (1+π¦) = π¦^2 (1+π₯) π₯^2+π₯^2 π¦ = π¦^2 + π¦^2 π₯ π₯^2 β π¦^2 = xy2 β x2y (π βπ) (π₯+π¦)=π₯π¦ (π¦ βπ₯) β(π βπ) (π₯+π¦)=π₯π¦ (π¦ βπ₯) β(π₯+π¦) = π₯π¦ βπ₯ βπ¦ = π₯π¦ βπ₯ = π₯π¦+π¦ βπ₯ = (π₯+1) π¦ π = (βπ)/(π + π) Differentiating π€.π.π‘.π₯. ππ¦/ππ₯ = π/ππ₯ ((βπ₯)/(π₯ + 1)) Using quotient rule As (π’/π£)β² = (π’^β² π£ β π£^β² π’)/π£^2 where u = βx & v = x + 1 ππ¦/ππ₯ = (π(βπ₯)/ππ₯ (π₯ + 1) β π(π₯ + 1)/ππ₯. (βπ₯))/(π₯ + 1)^2 ππ¦/ππ₯ = (β1 (π₯ + 1) + (1 + 0) π₯)/(π₯ + 1)^2 ππ¦/ππ₯ = (βπ₯ β 1 + π₯)/(π₯ + 1)^2 π π/π π = (βπ)/(π + π)^π