हिंदी

Differentiate the following w.r.t. x : sin-1 (1-25x21+25x2) - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Differentiate the following w.r.t. x : `sin^-1  ((1 - 25x^2)/(1 + 25x^2))`

योग

उत्तर

Let y = `sin^-1  ((1 - 25x^2)/(1 + 25x^2))`

= `sin^-1[(1 - (5x)^2)/(1 + (5x)^2)]`
Put 5x = tanθ.
Then θ = tan–1(5x)

∴ y = `sin^-1((1 - tan^2θ)/(1 + tan^2θ))`
= sin–1(cos2θ)

= `sin^-1[sin(pi/2 - 2θ)]`

= `pi/(2) - 2θ`

= `pi/(2) - 2tan^-1(5x)`
Differentiating w.r.t. x, we get
∴ `"dy"/"dx" = "d"/"dx"[pi/2 - 2tan^-1 (5x)]`

= `"d"/"dx"(pi/2) - 2"d"/"dx"[tan^-1(5x)]`

= `0 - 2 xx (1)/(1 + (5)^2)."d"/"dx"(5x)`

= `(-2)/(1 + 25x^2) xx 5`

= `(-10)/(1 + 25x^2)`.

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] 12 Standard HSC Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.2 | Q 9.09 | पृष्ठ ३०

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: `log[tan(x/2)]`


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


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


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


Differentiate the following w.r.t.x: `log[sec (e^(x^2))]`


Differentiate the following w.r.t.x: [log {log(logx)}]2


Differentiate the following w.r.t.x:

(x2 + 4x + 1)3 + (x3− 5x − 2)4 


Differentiate the following w.r.t.x:

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


Differentiate the following w.r.t.x: (1 + sin2 x)2 (1 + cos2 x)3 


Differentiate the following w.r.t.x:

log (sec 3x+ tan 3x)


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


Differentiate the following w.r.t.x: log[tan3x.sin4x.(x2 + 7)7]


Differentiate the following w.r.t.x:

y = (25)log5(secx) − (16)log4(tanx) 


Differentiate the following w.r.t. x :

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


Differentiate the following w.r.t. x : `"cosec"^-1((1)/(4cos^3 2x - 3cos2x))`


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((sin3x)/(1 + cos3x))`


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


Differentiate the following w.r.t. x : `cos^-1((sqrt(3)cosx - sinx)/(2))`


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


Differentiate the following w.r.t. x : cos–1(3x – 4x3)


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:

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


Differentiate the following w.r.t. x : `tan^-1((a + btanx)/(b - atanx))`


Differentiate the following w.r.t. x :

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


Differentiate the following w.r.t. x :

(sin x)tanx + (cos x)cotx 


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


If y = sin−1 (2x), find `("d"y)/(""d"x)` 


Differentiate `sin^-1((2cosx + 3sinx)/sqrt(13))` w.r. to x


If y = `sqrt(cos x + sqrt(cos x + sqrt(cos x + ...... ∞)`, show that `("d"y)/("d"x) = (sin x)/(1 - 2y)`


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


Derivative of (tanx)4 is ______ 


If f(x) = `(3x + 1)/(5x - 4)` and t = `(5 + 3x)/(x - 4)`, then f(t) is ______ 


If x2 + y2 - 2axy = 0, then `dy/dx` equals ______ 


Differentiate `tan^-1 (sqrt((3 - x)/(3 + x)))` w.r.t. x.


If y = log (sec x + tan x), find `dy/dx`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×