Advertisements
Advertisements
Question
If xy2 = 1, then prove that `2 "dy"/"dx" + y^3`= 0
Advertisements
Solution
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
RELATED QUESTIONS
Differentiate the following with respect to x.
3x4 – 2x3 + x + 8
Differentiate the following with respect to x.
`(x^2 + x + 1)/(x^2 - x + 1)`
Differentiate the following with respect to x.
`e^x/(1 + e^x)`
Differentiate the following with respect to x.
x3 ex
Differentiate the following with respect to x.
cos2 x
Differentiate the following with respect to x.
`sqrt(1 + x^2)`
Find `"dy"/"dx"` of the following function:
x = ct, y = `c/t`
Find y2 for the following function:
x = a cosθ, y = a sinθ
If = a cos mx + b sin mx, then show that y2 + m2y = 0.
If y = 2 sin x + 3 cos x, then show that y2 + y = 0.
