English

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

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

Let y = cos2[log(x2 + 7)]
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"{cos[log(x^2 + 7)]}^2`

= `2cos[log(x^2 + 7)]."d"/"dx"{cos[log(x^2 + 7)]}`

= `2cos[log(x^2 + 7)].{-sin[log(x^2 + 7)]}."d"/"dx"[log(x^2 + 7)]`

= `-2sin[log(x^2 + 7)].cos[log(x^2 + 7)] xx (1)/(x^2 + 7)."d"/"dx"(x^2 + 7)`

= `-sin[2log(x^2 + 7)] xx (1)/(x^2 + 7).(2x + 0)`  ...[∵ 2sinx · cosx = sin2x]

= `(-2x.sin[2log(x^2 + 7)])/(x^2 + 7)`.

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 + sqrt(x^2 + 1)`


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


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


Differentiate the following w.r.t.x: log[cos(x3 – 5)]


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 {log(logx)}]2


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


Differentiate the following w.r.t.x:

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


Differentiate the following w.r.t.x:

log (sec 3x+ tan 3x)


Differentiate the following w.r.t.x: `cot(logx/2) - log(cotx/2)`


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[(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(4x)


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


Differentiate the following w.r.t. x: 

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


Differentiate the following w. r. t. x.

cos–1(1 – x2)


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


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


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


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 :

`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 :

`(x +  1)^2/((x + 2)^3(x + 3)^4`


Differentiate the following w.r.t. x: `x^(tan^(-1)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 : x7.y5 = (x + y)12 


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


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


Differentiate sin2 (sin−1(x2)) 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 y = `sin^-1[("a"cosx - "b"sinx)/sqrt("a"^2 + "b"^2)]`, then find `("d"y)/("d"x)`


If f(x) = 3x - 2 and g(x) = x2, then (fog)(x) = ________.


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


y = {x(x - 3)}2 increases for all values of x lying in the interval.


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


If x = p sin θ, y = q cos θ, then `dy/dx` = ______ 


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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×