English

Differentiate the following w.r.t. x : x5.tan34xsin23x - Mathematics and Statistics

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

Let y = `(x^5.tan^3 4x)/(sin^2 3x)`

Then log y = `log[(x^5.tan^3 4x)/(sin^23x)]`

= logx5 + log tan34x – log sin23x

= 5logx + 3log (tan4x) – 2log (sin3x)

Differentiating both sides w.r.t. x, we get

`(1)/y."dy"/"dx" = 5"d"/"dx"(logx) + 3"d"/"dx"[log(tan4x)] - 2"d"/"dx"[log(sin3x)]`

= `5 xx (1)/x + 3 xx (1)/(tan4x)."d"/"dx"(tan4x) - 2 xx (1)/(sin3x)."d"/"dx"(sin3x)`

= `5/x + 3 xx (1)/(tan4x) xx sec^2  4x."d"/"dx"(4x) - 2 xx (1)/(sin3x) xx cos3x."d"/"dx"(3x)`

= `5/x + 3.(cos4x)/(sin4x) xx (1)/(cos^2 4x) xx 4 - 2cot3x xx 3`

= `5/x + (24)/(2sin4x.cos4x) - 6cot3x`

∴ `"dy"/"dx" = y[5/x + 24/(sin8x) - 6cot3x]`

= `(x^5.tan^3 4x)/(sin^2 3x)[5/x + 24"cosec"8x - 6cot3x]`.

shaalaa.com
  Is there an error in this question or solution?
Chapter 1: Differentiation - Exercise 1.3 [Page 39]

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(3x - 5) - 1/sqrt(3x - 5))^5`


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


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


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: `(1 + sinx°)/(1 - sinx°)`


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: `log(sqrt((1 - sinx)/(1 + sinx)))`


Differentiate the following w.r.t.x:

y = (25)log5(secx) − (16)log4(tanx) 


Differentiate the following w.r.t. x:

`(x^2 + 2)^4/(sqrt(x^2 + 5)`


Differentiate the following w.r.t. x : cot–1(x3)


Differentiate the following w.r.t. x :

cos3[cos–1(x3)]


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[(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 : `sin^-1((4sinx + 5cosx)/sqrt(41))`


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


Differentiate the following w.r.t. x :

`cos^-1[(3cos(e^x) + 2sin(e^x))/sqrt(13)]`


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  ((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((5 -x)/(6x^2 - 5x - 3))`


Differentiate the following w.r.t. x :

`(x +  1)^2/((x + 2)^3(x + 3)^4`


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


Differentiate the following w.r.t. x : (logx)x – (cos x)cotx 


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


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


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


Differentiate `cot^-1((cos x)/(1 + sinx))` w.r. to x


If x = `sqrt("a"^(sin^-1 "t")), "y" = sqrt("a"^(cos^-1 "t")), "then" "dy"/"dx"` = ______


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


Derivative of (tanx)4 is ______ 


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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×