हिंदी

Differentiate the following w.r.t.x: log[sec(ex2)]

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

Let y = `log[sec (e^(x^2))]`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"log[sec (e^(x^2))]`

= `(1)/(sec (e^(x^2))). "d"/"dx"[sec (e^(x^2))]`

= `(1)/(sec (e^(x^2))).sec(e^(x^2))tan(e^(x^2))."d"/"dx"(e^(x^2))`

= `tan(e^(x^2)).e^(x^2)."d"/"dx"(x^2)`

= `tan(e^(x^2)).e^(x^2).2x`
= `2x.e^(x^2)tan(e^(x^2))`.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Differentiation - Exercise 1.1 [पृष्ठ १२]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.1 | Q 2.15 | पृष्ठ १२

संबंधित प्रश्न

Differentiate the following w.r.t. x: `sqrt(x^2 + 4x - 7)`.


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: cos(x2 + a2)


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: `5^(sin^3x + 3)`


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


Differentiate the following w.r.t.x: sec[tan (x4 + 4)]


Differentiate the following w.r.t.x: `e^(log[(logx)^2 - logx^2]`


Differentiate the following w.r.t.x:

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


Differentiate the following w.r.t.x:

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


Differentiate the following w.r.t.x:

y = (25)log5(secx) − (16)log4(tanx) 


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


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


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(sqrt((1 + cosx)/2))`


Differentiate the following w.r.t.x:

tan–1 (cosec x + cot x)


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 : `sin^-1  ((1 - 25x^2)/(1 + 25x^2))`


Differentiate the following w.r.t. x :

`sin^(−1) ((1 − x^3)/(1 + x^3))`


Differentiate the following w.r.t. x :

`tan^(−1)[(2^(x + 2))/(1 − 3(4^x))]`


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


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


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


Differentiate the following w.r.t. x: xe + xx + ex + ee.


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


Differentiate the following w.r.t. x :

etanx + (logx)tanx 


Differentiate the following w.r.t. x : `10^(x^(x)) + x^(x(10)) + x^(10x)`


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


Differentiate y = `sqrt(x^2 + 5)` w.r. to x


Differentiate y = etanx w.r. to x


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


Differentiate sin2 (sin−1(x2)) w.r. to x


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


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


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


The volume of a spherical balloon is increasing at the rate of 10 cubic centimetre per minute. The rate of change of the surface of the balloon at the instant when its radius is 4 centimetres, is ______


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×