Advertisements
Advertisements
प्रश्न
Find `"dy"/"dx"` for the following function.
x2 – xy + y2 = 1
Advertisements
उत्तर
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
संबंधित प्रश्न
Differentiate the following with respect to x.
3x4 – 2x3 + x + 8
Differentiate the following with respect to x.
`e^x/(1 + x)`
Differentiate the following with respect to x.
cos2 x
Differentiate the following with respect to x.
cos3 x
Find `"dy"/"dx"` for the following function
xy – tan(xy)
If 4x + 3y = log(4x – 3y), then find `"dy"/"dx"`
Find `"dy"/"dx"` of the following function:
x = a(θ – sin θ), y = a(1 – cos θ)
Find y2 for the following function:
y = e3x+2
Find y2 for the following function:
y = log x + ax
If y = 2 sin x + 3 cos x, then show that y2 + y = 0.
