Advertisements
Advertisements
प्रश्न
Differentiate the following w.r.t. x : `cos^-1((e^x - e^(-x))/(e^x + e^(-x)))`
Advertisements
उत्तर
Let y = `cos^-1((e^x - e^(-x))/(e^x + e^(-x)))`
= `cos^-1[(e^x - 1/e^x)/(e^x + 1/e^x)]`
= `cos^-1((e^(2x) - 1)/(e^(2x) + 1))`
Put ex = tanθ.
Then θ = tan–1(ex)
∴ y = `cos^-1((tan^2θ - 1)/(tan^2θ + 1))`
= `cos^-1[-((1 - tan^2θ)/(1 + tan^2θ))]`
= cos–1(– cos2θ)
= cos–1[cos(π – 2θ)]
= π – 2θ
= π – 2tan–1(ex)
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"[pi - 2tan^-1(e^x)]`
= `"d"/"dx"(pi) - 2"d"/"dx"[tan^-1(e^x)]`
= `0 - 2 xx (1)/(1 + (e^x)^2)."d"/"dx"(e^x)`
= `(-2)/(1 + e^(2x)) xx e^x`
= `-(2e^x)/(1 + e^(2x)`
APPEARS IN
संबंधित प्रश्न
Differentiate the following w.r.t. x:
(x3 – 2x – 1)5
Differentiate the following w.r.t.x:
`(2x^(3/2) - 3x^(4/3) - 5)^(5/2)`
Differentiate the following w.r.t. x: `sqrt(x^2 + 4x - 7)`.
Differentiate the following w.r.t.x:
`(sqrt(3x - 5) - 1/sqrt(3x - 5))^5`
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: `(e^sqrt(x) + 1)/(e^sqrt(x) - 1)`
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[a^(cosx)/((x^2 - 3)^3 logx)]`
Differentiate the following w.r.t.x:
y = (25)log5(secx) − (16)log4(tanx)
Differentiate the following w.r.t. x : cosec–1 (e–x)
Differentiate the following w.r.t. x :
cos3[cos–1(x3)]
Differentiate the following w.r.t. x : `"cosec"^-1[1/cos(5^x)]`
Differentiate the following w.r.t. x : `cos^-1(sqrt((1 + cosx)/2))`
Differentiate the following w.r.t. x : `"cosec"^-1((1)/(4cos^3 2x - 3cos2x))`
Differentiate the following w.r.t. x : `tan^-1(sqrt((1 + cosx)/(1 - cosx)))`
Differentiate the following w.r.t.x:
tan–1 (cosec x + cot x)
Differentiate the following w.r.t. x :
`cot^-1[(sqrt(1 + sin ((4x)/3)) + sqrt(1 - sin ((4x)/3)))/(sqrt(1 + sin ((4x)/3)) - sqrt(1 - sin ((4x)/3)))]`
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 - 25x^2)/(1 + 25x^2))`
Differentiate the following w.r.t. x :
`tan^-1((5 -x)/(6x^2 - 5x - 3))`
Differentiate the following w.r.t. x :
`(x + 1)^2/((x + 2)^3(x + 3)^4`
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 : `(x^5.tan^3 4x)/(sin^2 3x)`
Differentiate the following w.r.t. x : (sin x)x
Differentiate the following w.r.t. x:
`x^(x^x) + e^(x^x)`
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`
Differentiate y = `sqrt(x^2 + 5)` w.r. to x
If f(x) is odd and differentiable, then f′(x) is
If y = `tan^-1[sqrt((1 + cos x)/(1 - cos x))]`, find `("d"y)/("d"x)`
If the function f(x) = `(log (1 + "ax") - log (1 - "bx))/x, x ≠ 0` is continuous at x = 0 then, f(0) = _____.
If x = `sqrt("a"^(sin^-1 "t")), "y" = sqrt("a"^(cos^-1 "t")), "then" "dy"/"dx"` = ______
If `t = v^2/3`, then `(-v/2 (df)/dt)` is equal to, (where f is acceleration) ______
Derivative of (tanx)4 is ______
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.
If y = cosec x0, then `"dy"/"dx"` = ______.
Find `(dy)/(dx)`, if x3 + x2y + xy2 + y3 = 81
If x = eθ, (sin θ – cos θ), y = eθ (sin θ + cos θ) then `dy/dx` at θ = `π/4` is ______.
If y = log (sec x + tan x), find `dy/dx`.
