हिंदी

Differentiate the following w.r.t. x : cot-1[1+sin (4x3)+1-sin (4x3)1+sin (4x3)-1-sin (4x3)] - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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)))]`

योग
Advertisements

उत्तर

Let y = `cot^-1[(sqrt(1 + sin  ((4x)/3)) + sqrt(1 - sin  ((4x)/3)))/(sqrt(1 + sin  ((4x)/3)) - sqrt(1 - sin  ((4x)/3)))]`

= `1 + sin ((4x)/3)`

= `1 + cos(pi/2 - (4x)/3)`

= `2cos^2(pi/4 - (2x)/3)`

∴ `sqrt(1 + sin((4x)/3)) = sqrt(2)cos(pi/4 - (2x)/3)`

Also, `1 - sin ((4x)/3)`

= `1 - cos(pi/2 - (4x)/3)`

= `2sin^2(pi/4 - (2x)/3)`

∴ `sqrt(1 - sin((4x)/3)) = sqrt(2)sin(pi/4 - (2x)/3)`

∴ `(sqrt(1 + sin  ((4x)/3)) + sqrt(1 - sin ((4x)/3)))/(sqrt(1 + sin((4x)/3) - sqrt(1 - sin((4x)/3)`

= `(sqrt(2)cos(pi/4 - (2x)/3) + sqrt(2)sin(pi/4 - (2x)/3))/(sqrt(2)cos(pi/4 - (2x)/3) - sqrt(2)sin(pi/4 - (2x)/3)`

= `(cos(pi/4 - (2x)/3) + sin(pi/4 - (2x)/3))/(cos(pi/4 - (2x)/3) - sin(pi/4 - (2x)/3)`

= `(1 + tan(pi/4 - (2x)/3))/(1 - tan(pi/4 - (2x)/3))                                ...["Dividing by" cos(pi/4 - (2x)/3)` 

= `(tan  pi/4 + tan(pi/4 - (2x)/3))/(1 - tan  pi/4. tan(pi/4 - (2x)/3))                        ...[∵ tan  pi/4 = 1]`

= `tan[pi/4 + pi/4 - (2x)/3]`

= `tan(pi/2 - (2x)/3)`

= `cot((2x)/3)`

∴ y = `cot^-1[cot((2x)/3)] = (2x)/(3)`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"((2x)/3)`

= `(2)/(3)"d"/"dx"(x)`

= `(2)/(3) xx 1`

= `(2)/(3)`

shaalaa.com
Differentiation
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Differentiation - Exercise 1.2 [पृष्ठ ३०]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.2 | Q 7.12 | पृष्ठ ३०

संबंधित प्रश्न

Differentiate the following w.r.t. x:

(x3 – 2x – 1)5


Differentiate the following w.r.t.x: cos(x2 + a2)


Differentiate the following w.r.t.x: cot3[log(x3)]


Differentiate the following w.r.t.x: `5^(sin^3x + 3)`


Differentiate the following w.r.t.x: `"cosec"(sqrt(cos x))`


Differentiate the following w.r.t.x: `sinsqrt(sinsqrt(x)`


Differentiate the following w.r.t.x:

sin2x2 – cos2x2 


Differentiate the following w.r.t.x: (1 + 4x)5 (3 + x −x2)


Differentiate the following w.r.t.x:

`(x^3 - 5)^5/(x^3 + 3)^3`


Differentiate the following w.r.t.x: `(1 + sinx°)/(1 - sinx°)`


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:

`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 : cot–1(x3)


Differentiate the following w.r.t. x :

`sin^-1(sqrt((1 + x^2)/2))`


Differentiate the following w.r.t. x :

`cos^-1(sqrt(1 - cos(x^2))/2)`


Differentiate the following w.r.t. x : `cot^-1((sin3x)/(1 + cos3x))`


Differentiate the following w.r.t. x : `cos^-1((3cos3x - 4sin3x)/5)`


Differentiate the following w.r.t. x : `tan^-1((2x)/(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 :

`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 : `tan^-1((8x)/(1 - 15x^2))`


Differentiate the following w.r.t.x:

`cot^-1((1 + 35x^2)/(2x))`


Differentiate the following w.r.t. x : `tan^-1((2^x)/(1 + 2^(2x + 1)))`


Differentiate the following w.r.t. x :

`tan^-1((5 -x)/(6x^2 - 5x - 3))`


Differentiate the following w.r.t. x : `(x^5.tan^3 4x)/(sin^2 3x)`


Differentiate the following w.r.t. x: `x^(tan^(-1)x`


Differentiate the following w.r.t. x : (sin x)x 


Differentiate the following w.r.t. x :

etanx + (logx)tanx 


Differentiate the following w.r.t. x : `[(tanx)^(tanx)]^(tanx) "at"  x = pi/(4)`


Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : x7.y5 = (x + y)12 


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: `log((x^20 - y^20)/(x^20 + y^20))` = 20


Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : `e^((x^7 - y^7)/(x^7 + y^7)` = a


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 sin2 (sin−1(x2)) w.r. to x


If `t = v^2/3`, then `(-v/2 (df)/dt)` is equal to, (where f is acceleration) ______ 


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 ______.


The volume of a spherical balloon is increasing at the rate of 10 cubic centimetre per minute. The rate of change of the surface of the balloon at the instant when its radius is 4 centimetres, is ______


Find `(dy)/(dx)`, if x3 + x2y + xy2 + y3 = 81


The value of `d/(dx)[tan^-1((a - x)/(1 + ax))]` is ______.


Let f(x) be a polynomial function of the second degree. If f(1) = f(–1) and a1, a2, a3 are in AP, then f’(a1), f’(a2), f’(a3) are in ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×