English

Differentiate the following w.r.t.x: coseccosec(cosx) - Mathematics and Statistics

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

Let y = `"cosec"(sqrt(cos x))`
Differentiating w.r.t. x,we get,
`"dy"/"dx" = "d"/"dx"["cosec"(sqrt(cos x))]`

= `-"cosec"(sqrt(cos x)).cot(sqrt(cos x))."d"/"dx"sqrt(cos x)`

= `-"cosec"(sqrt(cos x)).cot(sqrt(cos x)).(1)/(2sqrt(cos x))."d"/"dx"(cos x)`

= `-"cosec"(sqrt(cos x)).cot(sqrt(cos x)).(1)/(2sqrt(cos x)).(-sin x)`

= `(sin x. "cosec"(sqrt(cos x)).cot(sqrt(cos x)))/(2sqrt(cos x)`.

shaalaa.com
  Is there an error in this question or solution?
Chapter 1: Differentiation - Exercise 1.1 [Page 12]

RELATED QUESTIONS

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: log[cos(x3 – 5)]


Differentiate the following w.r.t.x: `e^(3sin^2x - 2cos^2x)`


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


Differentiate the following w.r.t.x:

log (sec 3x+ tan 3x)


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 : cosec–1 (e–x)


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 : `"cosec"^-1((1)/(4cos^3 2x - 3cos2x))`


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


Differentiate the following w.r.t. x : `tan^-1((cos7x)/(1 + sin7x))`


Differentiate the following w.r.t. x : `"cosec"^-1[(10)/(6sin(2^x) - 8cos(2^x))]`


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


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


Differentiate the following w.r.t.x:

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


Differentiate the following w.r.t. x : `tan^-1((2sqrt(x))/(1 + 3x))`


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: `x^(tan^(-1)x`


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 


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 : `e^((x^7 - y^7)/(x^7 + y^7)` = a


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 


If y is a function of x and log (x + y) = 2xy, then the value of y'(0) = ______.


Differentiate y = etanx w.r. to x


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


If f(x) is odd and differentiable, then f′(x) is


If y = `"e"^(1 + logx)` then 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"` = ______


A particle moves so that x = 2 + 27t - t3. The direction of motion reverses after moving a distance of ______ units.


The differential equation of the family of curves y = `"ae"^(2(x + "b"))` is ______.


If y = cosec x0, then `"dy"/"dx"` = ______.


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 ______


If x = eθ, (sin θ – cos θ), y = eθ (sin θ + cos θ) then `dy/dx` at θ = `π/4` is ______.


Diffierentiate: `tan^-1((a + b cos x)/(b - a cos x))` w.r.t.x.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×