मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Differentiate the following w.r.t. x : cos3[cos–1(x3)]

Advertisements
Advertisements

प्रश्न

Differentiate the following w.r.t. x :

cos3[cos–1(x3)]

बेरीज
Advertisements

उत्तर

Let y = cos3[cos–1(x3)]

Differentiating w.r.t. x, we get

`"dy"/"dx" = 3 cos^2 [cos^-1 (x^3)] . "d"/"dx"[cos(cos^-1 (x^3))]`

= `3 (x^3)^2 . "d"/"dx" [x^3]`

`"dy"/"dx" = 3(x^6) . 3(x^2)`

`"dy"/"dx" = 9x^8`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Differentiation - Exercise 1.2 [पृष्ठ २९]

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 + sqrt(x^2 + 1)`


Differentiate the following w.r.t.x:

`sqrt(e^((3x + 2) +  5)`


Differentiate the following w.r.t.x: `log[tan(x/2)]`


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


Differentiate the following w.r.t.x: cos2[log(x2 + 7)]


Differentiate the following w.r.t.x:

tan[cos(sinx)]


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


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


Differentiate the following w.r.t.x:

(x2 + 4x + 1)3 + (x3− 5x − 2)4 


Differentiate the following w.r.t.x: `x/(sqrt(7 - 3x)`


Differentiate the following w.r.t.x:

`sqrt(cosx) + sqrt(cossqrt(x)`


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 - cos3x)/(1 + cos3x)))`


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


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[a^(cosx)/((x^2 - 3)^3 logx)]`


Differentiate the following w.r.t. x : cosec–1 (e–x)


Differentiate the following w.r.t. x : `sin^4[sin^-1(sqrt(x))]`


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[(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((3cos3x - 4sin3x)/5)`


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 - x^2)/(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 : `cot^-1((1 - sqrt(x))/(1 + sqrt(x)))`


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


Differentiate the following w.r.t. x:

`x^(x^x) + e^(x^x)`


Differentiate the following w.r.t. x : `x^(e^x) + (logx)^(sinx)`


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 : `tan^-1((3x^2 - 4y^2)/(3x^2 + 4y^2))` = a2 


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 `cot^-1((cos x)/(1 + sinx))` 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)`


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)` = ______


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


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


Solve `x + y (dy)/(dx) = sec(x^2 + y^2)`


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×