Advertisements
Advertisements
प्रश्न
If y = tan x, then prove that y2 - 2yy1 = 0.
उत्तर
Given y = tan x ...(1)
Differentiating with respect to 'x' we get,
`y_1 = "dy"/"dx" = sec^2x` ....(2)
Differentiating again with respect to 'x' we get,
`y_2 = ("d"^2y)/"dx"^2 = 2 sec x "d"/"dx" (sec x)`
y2 = 2 sec x . sec x tan x
= 2 sec2x tan x
= 2y1y ...[using (1) and (2)]
⇒ y2 - 2yy1 = 0
Hence proved.
APPEARS IN
संबंधित प्रश्न
Differentiate the following with respect to x.
`(3 + 2x - x^2)/x`
Find `"dy"/"dx"` for the following function.
x2 – xy + y2 = 1
If 4x + 3y = log(4x – 3y), then find `"dy"/"dx"`
Differentiate the following with respect to x.
(sin x)x
Differentiate sin2x with respect to x2.
Find y2 for the following function:
y = log x + ax
Find y2 for the following function:
x = a cosθ, y = a sinθ
If y = 500e7x + 600e-7x, then show that y2 – 49y = 0.
If xy . yx , then prove that `"dy"/"dx" = y/x((x log y - y)/(y log x - x))`
If xy2 = 1, then prove that `2 "dy"/"dx" + y^3`= 0
