Advertisements
Advertisements
प्रश्न
Differentiate the following w.r.t. x : `"cosec"^-1((1)/(4cos^3 2x - 3cos2x))`
उत्तर
Let y = `"cosec"^-1((1)/(4cos^3 2x - 3cos2x))`
= `"cosec"^-1(1/(cos6x))` ...[∵ cos3x = 4cos3x – 3cosx]
= cosec–1(sec6x)
= `"cosec"^-1["cosec"(pi/2 - 6x)]`
= `pi/(2) - 6x`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"(pi/2 - 6x)`
= `"d"/"dx"(pi/2) - 6"d"/"dx"(x)`
= 0 – 6 x 1
= –6.
APPEARS IN
संबंधित प्रश्न
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: `log[sec (e^(x^2))]`
Differentiate the following w.r.t.x:
sin2x2 – cos2x2
Differentiate the following w.r.t.x:
(x2 + 4x + 1)3 + (x3− 5x − 2)4
Differentiate the following w.r.t.x: (1 + sin2 x)2 (1 + cos2 x)3
Differentiate the following w.r.t.x:
`sqrt(cosx) + sqrt(cossqrt(x)`
Differentiate the following w.r.t.x:
log (sec 3x+ tan 3x)
Differentiate the following w.r.t. x : cot–1(4x)
Differentiate the following w.r.t. x :
cos3[cos–1(x3)]
Differentiate the following w.r.t. x : `sin^4[sin^-1(sqrt(x))]`
Differentiate the following w.r.t. x : `cos^-1(sqrt((1 + cosx)/2))`
Differentiate the following w.r.t. x : `tan^-1[(1 + cos(x/3))/(sin(x/3))]`
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 : `cos^-1((sqrt(3)cosx - sinx)/(2))`
Differentiate the following w.r.t. x : cos–1(3x – 4x3)
Differentiate the following w.r.t. x : `cos^-1((e^x - e^(-x))/(e^x + e^(-x)))`
Differentiate the following w.r.t. x :
`sin^-1(4^(x + 1/2)/(1 + 2^(4x)))`
Differentiate the following w.r.t. x :
`sin^(−1) ((1 − x^3)/(1 + x^3))`
Differentiate the following w.r.t. x : `(x^2 + 3)^(3/2).sin^3 2x.2^(x^2)`
Differentiate the following w.r.t. x : (sin x)x
Differentiate the following w.r.t. x :
(sin x)tanx + (cos x)cotx
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
Solve the following :
The values of f(x), g(x), f'(x) and g'(x) are given in the following table :
| x | f(x) | g(x) | f'(x) | fg'(x) |
| – 1 | 3 | 2 | – 3 | 4 |
| 2 | 2 | – 1 | – 5 | – 4 |
Match the following :
| A Group – Function | B Group – Derivative |
| (A)`"d"/"dx"[f(g(x))]"at" x = -1` | 1. – 16 |
| (B)`"d"/"dx"[g(f(x) - 1)]"at" x = -1` | 2. 20 |
| (C)`"d"/"dx"[f(f(x) - 3)]"at" x = 2` | 3. – 20 |
| (D)`"d"/"dx"[g(g(x))]"at"x = 2` | 5. 12 |
Differentiate y = etanx w.r. to x
If y = `"e"^(1 + logx)` then find `("d"y)/("d"x)`
If y = `tan^-1[sqrt((1 + cos x)/(1 - cos x))]`, find `("d"y)/("d"x)`
Differentiate `sin^-1((2cosx + 3sinx)/sqrt(13))` w.r. to x
Differentiate `tan^-1((8x)/(1 - 15x^2))` w.r. to x
If x = `sqrt("a"^(sin^-1 "t")), "y" = sqrt("a"^(cos^-1 "t")), "then" "dy"/"dx"` = ______
Derivative of (tanx)4 is ______
y = {x(x - 3)}2 increases for all values of x lying in the interval.
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 = `(3x^2 - 4x + 7.5)^4, "then" dy/dx` is ______
The differential equation of the family of curves y = `"ae"^(2(x + "b"))` is ______.
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 ______.
Find `(dy)/(dx)`, if x3 + x3y + xy2 + y3 = 81
