Advertisements
Advertisements
प्रश्न
If y = `tan^-1[sqrt((1 + cos x)/(1 - cos x))]`, find `("d"y)/("d"x)`
उत्तर
`tan^-1[sqrt((1 + cos x)/(1 - cos x))]`
Put `1 + cos x = 2 cos^2 x/2`
`1 - cos x = 2 sin^2 x/2`
∵ `cos x = 2 cos^2 x/2 - 1 = 1 - 2 sin^2 x/2`
= `tan^-1[sqrt((2cos^2 (x/2))/(2sin^2 (x/2)))]`
= `tan^-1 [sqrt(cot^2(x/2))]`
= `tan^-1 [cot(x/2)]`
= `tan^-1 [tan(pi/2- x/2)] ...[cot theta = tan(pi/2 - theta)]`
= `pi/2 - x/2 ...[tan^-1 (tan theta) = theta]`
Differentiating w. r. t. x, we get
`("d"y)/("d"x) = "d"/("d"x) pi/2 - "d"/("d"x) x/2`
∴ `("d"y)/("d"x) = 0 - 1/2 "d"/("d"x) x`
∴ `("d"y)/("d"x) = -1/2`
APPEARS IN
संबंधित प्रश्न
Differentiate the following w.r.t.x: `sqrt(x^2 + 4x - 7)`
Differentiate the following w.r.t.x: `(8)/(3root(3)((2x^2 - 7x - 5)^11`
Differentiate the following w.r.t.x: `log[tan(x/2)]`
Differentiate the following w.r.t.x: `e^(3sin^2x - 2cos^2x)`
Differentiate the following w.r.t.x:
tan[cos(sinx)]
Differentiate the following w.r.t.x: sec[tan (x4 + 4)]
Differentiate the following w.r.t.x: (1 + 4x)5 (3 + x −x2)8
Differentiate the following w.r.t.x: `x/(sqrt(7 - 3x)`
Differentiate the following w.r.t.x:
`log(sqrt((1 + cos((5x)/2))/(1 - cos((5x)/2))))`
Differentiate the following w.r.t.x: `log[4^(2x)((x^2 + 5)/(sqrt(2x^3 - 4)))^(3/2)]`
Differentiate the following w.r.t.x: `log[(ex^2(5 - 4x)^(3/2))/root(3)(7 - 6x)]`
Differentiate the following w.r.t. x : `cot^-1[cot(e^(x^2))]`
Differentiate the following w.r.t. x : `"cosec"^-1[1/cos(5^x)]`
Differentiate the following w.r.t.x:
tan–1 (cosec x + cot x)
Differentiate the following w.r.t. x : `sin^-1((cossqrt(x) + sinsqrt(x))/sqrt(2))`
Differentiate the following w.r.t. x : `"cosec"^-1[(10)/(6sin(2^x) - 8cos(2^x))]`
Differentiate the following w.r.t. x : `sin^-1((1 - x^2)/(1 + x^2))`
Differentiate the following w.r.t. x : `sin^-1(2xsqrt(1 - x^2))`
Differentiate the following w.r.t. x :
`cos^-1 ((1 - 9^x))/((1 + 9^x)`
Differentiate the following w.r.t. x : `tan^-1((8x)/(1 - 15x^2))`
Differentiate the following w.r.t. x : `cot^-1((4 - x - 2x^2)/(3x + 2))`
Differentiate the following w.r.t. x : `((x^2 + 2x + 2)^(3/2))/((sqrt(x) + 3)^3(cosx)^x`
Differentiate the following w.r.t. x : `10^(x^(x)) + x^(x(10)) + x^(10x)`
Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : `sec((x^5 + y^5)/(x^5 - y^5))` = a2
Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : `tan^-1((3x^2 - 4y^2)/(3x^2 + 4y^2))` = a2
Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : `cos^-1((7x^4 + 5y^4)/(7x^4 - 5y^4)) = tan^-1a`
Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : `sin((x^3 - y^3)/(x^3 + y^3))` = a3
Differentiate y = `sqrt(x^2 + 5)` w.r. to x
Differentiate y = etanx w.r. to x
If x = `sqrt("a"^(sin^-1 "t")), "y" = sqrt("a"^(cos^-1 "t")), "then" "dy"/"dx"` = ______
If y = `1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + .....,` then `(d^2y)/(dx^2)` = ______
A particle moves so that x = 2 + 27t - t3. The direction of motion reverses after moving a distance of ______ units.
Let f(x) = `(1 - tan x)/(4x - pi), x ne pi/4, x ∈ [0, pi/2]`. If f(x) is continuous in `[0, pi/2]`, then f`(pi/4)` is ______.
Solve `x + y (dy)/(dx) = sec(x^2 + y^2)`
The value of `d/(dx)[tan^-1((a - x)/(1 + ax))]` is ______.
If `cos((x^2 - y^2)/(x^2 + y^2))` = log a, show that `dy/dx = y/x`
