Answer to Question #129368 in Calculus for Umar Nura

Question #129368
Find dy/dx at (1,2) of the function xy²-y-4x²+2=0
1
Expert's answer
2020-08-16T20:05:23-0400
"xy^2-y-4x^2+2=0"

Differentiate both sides with respect to "x"


"{d\\over dx}(xy^2-y-4x^2+2)={d\\over dx}(0)"

Use the Chain rule


"y^2+x(2y){dy\\over dx}-{dy\\over dx}-8x+0=0"

Solve for "\\dfrac{dy}{dx}"

"\\dfrac{dy}{dx}=\\dfrac{8x-y^2}{2xy-1}"

Substitute "x=1, y=2"


"\\dfrac{dy}{dx}\\big|_{(1,2)}=\\dfrac{8(1)-(2)^2}{2(1)(2)-1}=\\dfrac{4}{3}"


"\\dfrac{dy}{dx}\\big|_{(1,2)}=\\dfrac{4}{3}"




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