Advertisements
Advertisements
प्रश्न
Find `dx/dy` in the following:
`y = cos^(-1) ((2x)/(1+x^2)), -1 < x < 1`
उत्तर
y `= cos^-1 ((2x)/(1 + x^2))`
Let, x `= tan theta => theta = tan^-1 x`
`therefore y = cos^-1 ((2 tan theta)/(1 + tan^2 theta))`
`= cos^-1 (sin 2 theta)`
`= cos^-1 {cos (pi/2 - 2 theta)}`
`y = pi/2 - 2 theta`
`y = pi/2 = 2 tan^-1 x`
`therefore dy/dx = 0 - 2x 1/(1 + x^2)`
`⇒ dy/dx= - 2/(1 + x^2)`
APPEARS IN
संबंधित प्रश्न
Differentiate `cos^-1((3cosx-2sinx)/sqrt13)` w. r. t. x.
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 : ` d/dx cos^−1 ((x−x^(−1))/(x+x^(−1)))`
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 = cos^(-1) ((1-x^2)/(1+x^2)), 0 < x < 1`
Find `dy/dx` in the following:
`y = sin^(-1) ((1-x^2)/(1+x^2)), 0 < x < 1`
Find `dy/dx` in the following:
`y = sec^(-1) (1/(2x^2 - 1)), 0 < x < 1/sqrt2`
Differentiate w.r.t. x the function:
`cot^(-1) [(sqrt(1+sinx) + sqrt(1-sinx))/(sqrt(1+sinx) - sqrt(1-sinx))]`, ` 0 < x < pi/2`
If `xsqrt(1+y) + y sqrt(1+x) = 0`, for, −1 < x <1, prove that `dy/dx = 1/(1+ x)^2`
Find the approximate value of tan−1 (1.001).
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 y = cos (sin x), show that: `("d"^2"y")/("dx"^2) + "tan x" "dy"/"dx" + "y" "cos"^2"x" = 0`
If `log (x^2 + y^2) = 2 tan^-1 (y/x)`, show that `(dy)/(dx) = (x + y)/(x - y)`
If `"y" = (sin^-1 "x")^2, "prove that" (1 - "x"^2) (d^2"y")/(d"x"^2) - "x" (d"y")/(d"x") - 2 = 0`.
If y = `(sin^-1 x)^2,` prove that `(1-x^2) (d^2y)/dx^2 - x dy/dx -2 = 0.`
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 ____________.
`"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 ____________.
The derivative of `sin^-1 ((2x)/(1 + x^2))` with respect to `cos^-1 [(1 - x^2)/(1 + x^2)]` is equal to
Let f(x) = `cos(2tan^-1sin(cot^-1sqrt((1 - x)/x))), 0 < x < 1`. Then ______.
Differentiate `sec^-1 (1/sqrt(1 - x^2))` w.r.t. `sin^-1 (2xsqrt(1 - x^2))`.
