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

Differentiate the following w.r.t. x : at[(tanx)tanx]tanxat x=π4 - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Differentiate the following w.r.t. x : `[(tanx)^(tanx)]^(tanx) "at"  x = pi/(4)`

बेरीज
Advertisements

उत्तर

Let y = `[(tanx)^(tanx)]^(tanx)`
∴ log y = `log[(tanx)^(tanx)]tanx`
= tanx. log(tanx)tanx
= tanx. tanx log(tan x)
= (tanx)2. log(tan x)
Differentiating both sides w.r.t. x, we get
`1/y."dy"/"dx" = "d"/"dx"[tanx)^2.log(tanx)]`

= `(tanx)^2."d"/"dx"(log tanx) + (log tanx)."d"/"dx"(tanx)^2`

= `(tanx)^2. xx 1/tanx."d"/"dx"(tanx) + (log tanx) xx 2tanx."d"/"dx"(tanx)`

= `(tanx)^2 xx 1/tanx.sec^2x + (log tanx) xx 2 tanxsec^2x`

∴ `"dy"/"dx" = y[(tanx)(sec^2x) + (logtanx)(2tanxsec^2x)]`

= [(tanx)tanx]tanx.(tanxsec2x)[1 + 2logtanx]

If x = `pi/(4)`, then

`"dy"/"dx" = [(tan pi/4)^(tan  pi/4)]^(tan  pi/4)(tan  pi/4 sec^2  pi/4)[1 + 2log tan  pi/4]`
= `[(1)^1]^1.[1(sqrt(2))^2][1 + 2log1]`
= 1 x 2 x 1                  ...[∵ log 1 = 0]
= 2.

shaalaa.com
Differentiation
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Differentiation - Exercise 1.3 [पृष्ठ ४०]

APPEARS IN

संबंधित प्रश्‍न

Differentiate the following w.r.t.x:

`sqrt(x^2 + sqrt(x^2 + 1)`


Differentiate the following w.r.t.x: `(8)/(3root(3)((2x^2 - 7x - 5)^11`


Differentiate the following w.r.t.x:

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


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: `sqrt(tansqrt(x)`


Differentiate the following w.r.t.x: cot3[log(x3)]


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:

(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:

`(x^3 - 5)^5/(x^3 + 3)^3`


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


Differentiate the following w.r.t.x:

log (sec 3x+ tan 3x)


Differentiate the following w.r.t.x:

`(e^(2x) - e^(-2x))/(e^(2x) + e^(-2x))`


Differentiate the following w.r.t.x: log[tan3x.sin4x.(x2 + 7)7]


Differentiate the following w.r.t.x:

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


Differentiate the following w.r.t.x:

`log(sqrt((1 + cos((5x)/2))/(1 - cos((5x)/2))))`


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


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


Differentiate the following w.r.t. x :

cos3[cos–1(x3)]


Differentiate the following w.r.t. x : `"cosec"^-1[1/cos(5^x)]`


Differentiate the following w.r.t. x : `tan^-1[(1 + cos(x/3))/(sin(x/3))]`


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


Differentiate the following w.r.t. x : cos–1(3x – 4x3)


Differentiate the following w.r.t. x :

`cos^-1  ((1 - 9^x))/((1 + 9^x)`


Differentiate the following w.r.t. x : `cot^-1((4 - x - 2x^2)/(3x + 2))`


Differentiate the following w.r.t. x :

`(x +  1)^2/((x + 2)^3(x + 3)^4`


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


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


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 : `x^(e^x) + (logx)^(sinx)`


Differentiate y = etanx w.r. to x


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


Differentiate `tan^-1((8x)/(1 - 15x^2))` 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 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"` = ______


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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×