Advertisements
Advertisements
प्रश्न
If y = tan x, then prove that y2 - 2yy1 = 0.
Advertisements
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.
x sin x
Differentiate the following with respect to x.
ex sin x
Differentiate the following with respect to x.
sin x cos x
Differentiate the following with respect to x.
cos3 x
Differentiate the following with respect to x.
sin(x2)
Differentiate the following with respect to x.
xsin x
If y = 2 + log x, then show that xy2 + y1 = 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
If y = 2 sin x + 3 cos x, then show that y2 + y = 0.
