Advertisements
Advertisements
प्रश्न
If y = `(sin^-1 x)^2,` prove that `(1-x^2) (d^2y)/dx^2 - x dy/dx -2 = 0.`
उत्तर
Here,
`y = (sin^-1 x)^2`
⇒ `y_1 = 2 sin^-1 x 1/sqrt(1-x^2)`
⇒ `y_2 = 2/(1-x^2) + (2x sin^-1 x)/(1-x^2)^(3/2)`
⇒ `y_2 = 2/(1-x^2) + (2x sin^-1 x)/((1-x^2)sqrt(1-x^2)`
⇒ `y_2 = 2/(1-x^2) + (xy_1)/((1-x^2)`
⇒ `y_2 (1-x^2) = 2 + xy_1`
⇒ `y_2 (1-x^2) - xy_1 - 2 =0`
⇒ Therefore, `(1 -x^2) (d^2y)/dx^2 - x dy/dx - 2 = 0`
Hence proved.
APPEARS IN
संबंधित प्रश्न
If `sec((x+y)/(x-y))=a^2. " then " (d^2y)/dx^2=........`
(a) y
(b) x
(c) y/x
(d) 0
If `y=sin^-1(3x)+sec^-1(1/(3x)), ` find dy/dx
Differentiate `tan^(-1)(sqrt(1-x^2)/x)` with respect to `cos^(-1)(2xsqrt(1-x^2))` ,when `x!=0`
Find the derivative of the following function f(x) w.r.t. x, at x = 1 :
`f(x)=cos^-1[sin sqrt((1+x)/2)]+x^x`
if `y = sin^(-1)[(6x-4sqrt(1-4x^2))/5]` Find `dy/dx `.
If `y=tan^(−1) ((sqrt(1+x^2)+sqrt(1−x^2))/(sqrt(1+x^2)−sqrt(1−x^2)))` , x2≤1, then find dy/dx.
Find `dy/dx` in the following:
`y = tan^(-1) ((3x -x^3)/(1 - 3x^2)), - 1/sqrt3 < x < 1/sqrt3`
Find `dy/dx` in the following:
`y = sin^(-1) ((1-x^2)/(1+x^2)), 0 < x < 1`
Find `dx/dy` in the following:
`y = cos^(-1) ((2x)/(1+x^2)), -1 < x < 1`
Find `dy/dx` in the following:
`y = sin^(-1)(2xsqrt(1-x^2)), -1/sqrt2 < x < 1/sqrt2`
Find `dy/dx` in the following:
`y = sec^(-1) (1/(2x^2 - 1)), 0 < x < 1/sqrt2`
Find `dy/dx, if y = sin^-1 x + sin^-1 sqrt (1 - x^2) , 0<x <1`
If `sqrt(1-x^2) + sqrt(1- y^2)` = a(x − y), show that dy/dx = `sqrt((1-y^2)/(1-x^2))`
if `x = tan(1/a log y)`, prove that `(1+x^2) (d^2y)/(dx^2) + (2x + a) (dy)/(dx) = 0`
Solve `cos^(-1)(sin cos^(-1)x) = pi/2`
If y = (sec-1 x )2 , x > 0, show that
`x^2 (x^2 - 1) (d^2 y)/(dx^2) + (2x^3 - x ) dy/dx -2 = 0`
If `log (x^2 + y^2) = 2 tan^-1 (y/x)`, show that `(dy)/(dx) = (x + y)/(x - y)`
The function f(x) = cot x is discontinuous on the set ______.
Trigonometric and inverse-trigonometric functions are differentiable in their respective domain.
`lim_("h" -> 0) (1/("h"^2 sqrt(8 + "h")) - 1/(2"h"))` is equal to ____________.
`lim_("x" -> -3) sqrt("x"^2 + 7 - 4)/("x" + 3)` is equal to ____________.
`"d"/"dx" {"cosec"^-1 ((1 + "x"^2)/(2"x"))}` is equal to ____________.
If `"y = sin"^-1 ((sqrt"x" - 1)/(sqrt"x" + 1)) + "sec"^-1 ((sqrt"x" + 1)/(sqrt"x" - 1)), "x" > 0, "then" "dy"/"dx"` is ____________.
If y `= "cos"^2 ((3"x")/2) - "sin"^2 ((3"x")/2), "then" ("d"^2"y")/("dx"^2)` is ____________.
The derivative of sin x with respect to log x is ____________.
If y = sin–1x, then (1 – x2)y2 is equal to ______.
