Advertisements
Advertisements
प्रश्न
If xy2 = 1, then prove that `2 "dy"/"dx" + y^3`= 0
Advertisements
उत्तर
Given xy2 = 1 ....(1)
Differentiating with respect to 'x' we get,
`x*2y "dy"/"dx" + y^2 (1) = 0` ...[using product rule]
Multiplying by y throughout we get,
`2xy^2 "dy"/"dx" + y^3` = 0
`=> 2(1) * "dy"/"dx" + y^3` = 0
`=> 2(1) "dy"/"dx" + y^3 = 0` ...[using (1)]
`=> 2 "dy"/"dx" + y^3` = 0
Hence proved.
APPEARS IN
संबंधित प्रश्न
Differentiate the following with respect to x.
`5/x^4 - 2/x^3 + 5/x`
Differentiate the following with respect to x.
`(x^2 + x + 1)/(x^2 - x + 1)`
Differentiate the following with respect to x.
ex sin x
Differentiate the following with respect to x.
`sqrt(1 + x^2)`
Find `"dy"/"dx"` for the following function.
x2 – xy + y2 = 1
Differentiate the following with respect to x.
(sin x)x
Differentiate the following with respect to x.
(sin x)tan x
If y = 2 + log x, then show that xy2 + y1 = 0.
If y = tan x, then prove that y2 - 2yy1 = 0.
If y = 2 sin x + 3 cos x, then show that y2 + y = 0.
