मराठी

Trigonometric and inverse - trigonometric functions are differentiable in their respective domain. - Mathematics

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प्रश्न

Trigonometric and inverse-trigonometric functions are differentiable in their respective domain.

पर्याय

  • True

  • False

MCQ
चूक किंवा बरोबर

उत्तर

This statement is True.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Continuity And Differentiability - Exercise [पृष्ठ ११६]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 5 Continuity And Differentiability
Exercise | Q 105 | पृष्ठ ११६

संबंधित प्रश्‍न

If `sec((x+y)/(x-y))=a^2. " then " (d^2y)/dx^2=........`

(a) y

(b) x

(c) y/x

(d) 0


If `y=sin^-1(3x)+sec^-1(1/(3x)), `  find dy/dx


Differentiate `tan^(-1)(sqrt(1-x^2)/x)` with respect to `cos^(-1)(2xsqrt(1-x^2))` ,when `x!=0`


Find : ` d/dx cos^−1 ((x−x^(−1))/(x+x^(−1)))`


Find the derivative of the following function f(x) w.r.t. x, at x = 1 : 

`f(x)=cos^-1[sin sqrt((1+x)/2)]+x^x`


if `y = sin^(-1)[(6x-4sqrt(1-4x^2))/5]` Find `dy/dx `.


If `y=tan^(−1) ((sqrt(1+x^2)+sqrt(1−x^2))/(sqrt(1+x^2)−sqrt(1−x^2)))` , x21, then find dy/dx.


Find `dy/dx` in the following:

`y = cos^(-1) ((1-x^2)/(1+x^2)), 0 < x < 1`


Find `dy/dx` in the following:

`y = sin^(-1) ((1-x^2)/(1+x^2)), 0 < x < 1`


Find `dx/dy` in the following:

`y = cos^(-1) ((2x)/(1+x^2)), -1 < x < 1`


Find `dy/dx` in the following:

`y = sec^(-1) (1/(2x^2 - 1)), 0 < x < 1/sqrt2`


If `sqrt(1-x^2)  + sqrt(1- y^2)` =  a(x − y), show that dy/dx = `sqrt((1-y^2)/(1-x^2))`


if `x = tan(1/a log y)`, prove that `(1+x^2) (d^2y)/(dx^2) + (2x + a) (dy)/(dx) = 0`


Find \[\frac{dy}{dx}\] at \[t = \frac{2\pi}{3}\] when x = 10 (t – sin t) and y = 12 (1 – cos t).


If y = (sec-1 x )2 , x > 0, show that 

`x^2 (x^2 - 1) (d^2 y)/(dx^2) + (2x^3 - x ) dy/dx -2 = 0`


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`lim_("x"-> 0) ("cosec x - cot x")/"x"`  is equal to ____________.


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The derivative of `sin^-1 ((2x)/(1 + x^2))` with respect to `cos^-1 [(1 - x^2)/(1 + x^2)]` is equal to


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