Advertisements
Advertisements
Question
Find `"dy"/"dx"` for the following function.
x2 – xy + y2 = 1
Advertisements
Solution
x2 – xy + y2 = 1
Differentiating both side with respect to x,
`"d"/"dx" (x^2) - "d"/"dx" (xy) + "d"/"dx" (y^2) = "d"/"dx"(7)`
`2x - [x "d"/"dx" (y) + y "d"/"dx" (x)] + 2y "dy"/"dx" = 0`
`2x - [x "dy"/"dx" + y * 1] + 2y "dy"/"dx" = 0`
`2x - x"dy"/"dx" - y + 2y "dy"/"dx"` = 0
`"dy"/"dx" [2x - x] = y - 2x`
`"dy"/"dx" = (y - 2x)/(2y - x)`
APPEARS IN
RELATED QUESTIONS
Differentiate the following with respect to x.
3x4 – 2x3 + x + 8
Differentiate the following with respect to x.
`e^x/(1 + e^x)`
Differentiate the following with respect to x.
sin2 x
Differentiate the following with respect to x.
cos3 x
Differentiate the following with respect to x.
sin(x2)
If `xsqrt(1 + y) + ysqrt(1 + x)` = 0 and x ≠ y, then prove that `"dy"/"dx" = - 1/(x + 1)^2`
Differentiate the following with respect to x.
(sin x)tan x
If y = 2 + log x, then show that xy2 + y1 = 0.
If y = `(x + sqrt(1 + x^2))^m`, then show that (1 + x2) y2 + xy1 – m2y = 0
If xy2 = 1, then prove that `2 "dy"/"dx" + y^3`= 0
