English

Aaaatan-1(3a2x-x3a3-3ax2),-13<xa<13 - Mathematics

Advertisements
Advertisements

Question

`tan^-1 ((3"a"^2x - x^3)/("a"^3 - 3"a"x^2)), (-1)/sqrt(3) < x/"a" < 1/sqrt(3)`

Sum
Advertisements

Solution

Let y = `tan^-1 [(3"a"^2x - x^3)/("a"^3 - 3"a"x^2)]`

Put x = a tan θ

∴ θ = `tan^-1  x/"a"`

y = `tan^-1 [(3"a"^2 * "a"tantheta - "a"^3 tan^3 theta)/("a"^3 - 3"a"*"a"^2 tan^2theta)]`

⇒ y = `tan^-1 [(3"a"^2 tantheta - "a"^3 tan^3theta)/("a"^3 - 3"a"^3 tan^2theta)]`

⇒ y = `tan^-1 [(3tan theta - tan^2ttheta)/(1 - 3tan^2 theta)]`

⇒ y = `tan^-1 [tan 3theta)]`  .......`[because tan 3theta = (3tantheta - tan^2theta)/(1 - 3tan^2theta)]`

⇒ y = 3θ

⇒ y = `3tan^-1  x/"a"`

Differentiating both sides w.r.t. x

`"dy"/"dx" = 3*"d"/"dx" (tan^-1  x/"a")`

= `3* 1/(1 + x^2/"a"^2) * "d"/"dx" * (x/"a")`

= `3 * "a"^2/("a"^2 + x^2) * 1/"a"`

= `(3"a")/("a"^2 + x^2)`

Hence, `"dy"/"dx" = (3"a")/("a"^2 + x^2)`.

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Continuity And Differentiability - Exercise [Page 110]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 5 Continuity And Differentiability
Exercise | Q 42 | Page 110

RELATED QUESTIONS

Differentiate the function with respect to x.

sin (ax + b)


Differentiate the function with respect to x.

`sec(tan (sqrtx))`


Differentiate the function with respect to x. 

cos x3 . sin2 (x5)


Differentiate the function with respect to x. 

`2sqrt(cot(x^2))`


Differentiate the function with respect to x:

(3x2 – 9x + 5)9


Differentiate the function with respect to x:

sin3 x + cos6 x


Differentiate the function with respect to x:

`sin^(–1)(xsqrtx), 0 ≤ x ≤ 1`


Find `dy/dx`, if y = 12 (1 – cos t), x = 10 (t – sin t), `-pi/2 < t < pi/2`.


Does there exist a function which is continuos everywhere but not differentiable at exactly two points? Justify your answer?


`"If y" = (sec^-1 "x")^2 , "x" > 0  "show that"  "x"^2 ("x"^2 - 1) (d^2"y")/(d"x"^2) + (2"x"^3 - "x") (d"y")/(d"x") - 2 = 0`


If f(x) = x + 1, find `d/dx (fof) (x)`


Let f(x) = x|x|, for all x ∈ R. Discuss the derivability of f(x) at x = 0


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


If y = tanx + secx, prove that `("d"^2y)/("d"x^2) = cosx/(1 - sinx)^2`


Differentiate `tan^-1 (sqrt(1 - x^2)/x)` with respect to`cos^-1(2xsqrt(1 - x^2))`, where `x ∈ (1/sqrt(2), 1)`


Differential coefficient of sec (tan–1x) w.r.t. x is ______.


If u = `sin^-1 ((2x)/(1 + x^2))` and v = `tan^-1 ((2x)/(1 - x^2))`, then `"du"/"dv"` is ______.


|sinx| is a differentiable function for every value of x.


cos |x| is differentiable everywhere.


`sin sqrt(x) + cos^2 sqrt(x)`


(sin x)cosx 


(x + 1)2(x + 2)3(x + 3)4


`tan^-1 (("a"cosx - "b"sinx)/("b"cosx - "a"sinx)), - pi/2 < x < pi/2` and `"a"/"b" tan x > -1`


`tan^-1 ((sqrt(1 + x^2) + sqrt(1 - x^2))/(sqrt(1 + x^2) - sqrt(1 - x^2))), -1 < x < 1, x ≠ 0`


For the curve `sqrt(x) + sqrt(y)` = 1, `"dy"/"dx"` at `(1/4, 1/4)` is ______.


The differential coefficient of `"tan"^-1 ((sqrt(1 + "x") - sqrt (1 - "x"))/(sqrt (1+ "x") + sqrt (1 - "x")))` is ____________.


A function is said to be continuous for x ∈ R, if ____________.


The rate of increase of bacteria in a certain culture is proportional to the number present. If it doubles in 5 hours then in 25 hours, its number would be


`d/(dx)[sin^-1(xsqrt(1 - x) - sqrt(x)sqrt(1 - x^2))]` is equal to


If `ysqrt(1 - x^2) + xsqrt(1 - y^2)` = 1, then prove that `(dy)/(dx) = - sqrt((1 - y^2)/(1 - x^2))`


Let f: R→R and f be a differentiable function such that f(x + 2y) = f(x) + 4f(y) + 2y(2x – 1) ∀ x, y ∈ R and f’(0) = 1, then f(3) + f’(3) is ______.


If f(x) = `{{:((sin(p  +  1)x  +  sinx)/x,",", x < 0),(q,",", x = 0),((sqrt(x  +  x^2)  -  sqrt(x))/(x^(3//2)),",", x > 0):}`

is continuous at x = 0, then the ordered pair (p, q) is equal to ______.


If f(x) = `{{:(ax + b; 0 < x ≤ 1),(2x^2 - x; 1 < x < 2):}` is a differentiable function in (0, 2), then find the values of a and b.


If f(x) = `{{:(x^2"," if x ≥ 1),(x"," if x < 1):}`, then show that f is not differentiable at x = 1.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×