English

2cos2x

Advertisements
Advertisements

Question

`2^(cos^(2_x)`

Sum
Advertisements

Solution

Let y = `2^(cos^(2_x)`

Taking log on both sides, we get

log y = `log 2^(cos^(2_x)`

⇒ log y = `cos^2x * log 2`

Differentiating both sides w.r.t. x

⇒ `1/y * "dy"/"dx" = log 2* "d"/"dx" cos^2x`

⇒ `1/y * "dy"/"dx" = log 2 [2 cos x * "d"/"dx" cos x]`

⇒ `1/y * "dy"/"dx" = log 2 [2 cos x(-sin x)]`

⇒ `1/y * "dy"/"dx" = log 2 (- sin 2x)`

`"dy"/"dx" = - y * log 2 sin 2x`

Hence, `"dy"/"dx" = -2^(cos^2x) (log 2 sin 2x)`

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Continuity And Differentiability - Exercise [Page 109]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 5 Continuity And Differentiability
Exercise | Q 25 | Page 109

RELATED QUESTIONS

 

If `y=log[x+sqrt(x^2+a^2)]` show that `(x^2+a^2)(d^2y)/(dx^2)+xdy/dx=0`

 

Differentiate the function with respect to x. 

cos x . cos 2x . cos 3x


Differentiate the function with respect to x.

`x^(xcosx) + (x^2 + 1)/(x^2 -1)`


Find `bb(dy/dx)` for the given function:

yx = xy


Find `bb(dy/dx)` for the given function:

xy = `e^((x - y))`


Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned below:

  1. By using the product rule.
  2. By expanding the product to obtain a single polynomial.
  3. By logarithmic differentiation.

Do they all give the same answer?


If u, v and w are functions of x, then show that `d/dx(u.v.w) = (du)/dx v.w + u. (dv)/dx.w + u.v. (dw)/dx` in two ways-first by repeated application of product rule, second by logarithmic differentiation.


Differentiate the function with respect to x:

xx + xa + ax + aa, for some fixed a > 0 and x > 0


Find `dy/dx` if y = x+ 5x


Find `(d^2y)/(dx^2)` , if y = log x


Find `"dy"/"dx"` , if `"y" = "x"^("e"^"x")`


Differentiate : log (1 + x2)  w.r.t. cot-1 x. 


If `(sin "x")^"y" = "x" + "y", "find" (d"y")/(d"x")`


If y = (log x)x + xlog x, find `"dy"/"dx".`


If log (x + y) = log(xy) + p, where p is a constant, then prove that `"dy"/"dx" = (-y^2)/(x^2)`.


If xy = ex–y, then show that `"dy"/"dx" = logx/(1 + logx)^2`.


`"If"  y = sqrt(logx + sqrt(log x + sqrt(log x + ... ∞))), "then show that"  dy/dx = (1)/(x(2y - 1).`


If ey = yx, then show that `"dy"/"dx" = (logy)^2/(log y - 1)`.


If x = `(2bt)/(1 + t^2), y = a((1 - t^2)/(1 + t^2)), "show that" "dx"/"dy" = -(b^2y)/(a^2x)`.


If y = A cos (log x) + B sin (log x), show that x2y2 + xy1 + y = 0.


If y = log [cos(x5)] then find `("d"y)/("d"x)`


If y = `log[4^(2x)((x^2 + 5)/sqrt(2x^3 - 4))^(3/2)]`, find `("d"y)/("d"x)`


If log5 `((x^4 + "y"^4)/(x^4 - "y"^4))` = 2, show that `("dy")/("d"x) = (12x^3)/(13"y"^2)`


lf y = `2^(x^(2^(x^(...∞))))`, then x(1 - y logx logy)`dy/dx` = ______  


`"d"/"dx" [(cos x)^(log x)]` = ______.


`log [log(logx^5)]`


`lim_("x" -> 0)(1 - "cos x")/"x"^2` is equal to ____________.


If y = `(1 + 1/x)^x` then `(2sqrt(y_2(2) + 1/8))/((log  3/2 - 1/3))` is equal to ______.


Find `dy/dx`, if y = (sin x)tan x – xlog x.


If \[y=x^x+x^{\frac{1}{x}}\] then \[\frac{\mathrm{d}y}{\mathrm{d}x}\] is equal to


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×