Advertisements
Advertisements
Question
`tan^-1 (secx + tanx), - pi/2 < x < pi/2`
Advertisements
Solution
Let y = tan–1(sec x + tan x)
Differentiating both sides w.r.t. x
`"dy"/"dx" = "d"/"dx" [tan^-1 (secx + tanx)]`
= `1/(1 + (secx + tanx)^2) * "d"/"dx"(secx + tanx)`
= `1/(1 + sec^2 + tan^2x + 2 sec x tanx) * (secx tanx + sec^2x)`
= `1/((1 + tan^2x) + sec^2x + 2secx tanx) * secx(tanx + secx)`
= `1/(sec^2x + sec^2x + 2secx tanx) * secx(tanx + secx)`
= `1/(2sec^2x + 2secx tanx) * secx(tanx + secx)`
= `1/(2secx(secx + tanx)) * secx(tanx + secx)`
= `1/2`
Hence, `"dy"/"dx" = 1/2`
Alternative solution:
Let y = `tan^-1 (secx + tanx), (-pi)/2 < x < pi/2`
= `tan^-1 (1/cosx + sinx/cosx)`
= `tan^-1 ((1 + sinx)/cosx)`
= `tan^-1 [(cos^2 x/2 + sin^2 x/2 + 2sin x/2 cos x/2)/(cos^2 x/2 - sin^2 x/2)]` ......`[(because 2x = 2sinx cosx),(cos2x = cos^2x - sin^2x)]`
= `tan^-1 [(cos x/2 + sin x/2)^2/((cos x/2 + sin x/2)(cos x/2 - sin x/2))]`
= `tan^-1 [(cos x/2 + sin x/2)/(cos x/2 - sin x/2)]`
= `tan^-1 [(1 + tan x/2)/(1 - tan x/2)]` .....[Dividing the Nr. and Den. by cos `x/2`]
= `tan^-1 [(tan pi/4 + tan x/2),(1 - tan pi/4 * tan x/2)]`
= `tan^-1 [tan (pi/4 + x/2)]`
∴ y = `pi/4 + x/2`
Differentiating both sides w.r.t. x
`"dy"/"dx" = 1/2 "d"/"dx" (x)`
= `1/2 * 1`
= `1/2`
Hence, `"dy"/"dx" = 1/2`.
APPEARS IN
RELATED QUESTIONS
Differentiate the function with respect to x.
cos (sin x)
Differentiate the function with respect to x.
`sec(tan (sqrtx))`
Differentiate the function with respect to x.
`(sin (ax + b))/cos (cx + d)`
Differentiate the function with respect to x.
`cos (sqrtx)`
Differentiate the function with respect to x:
(3x2 – 9x + 5)9
Differentiate the function with respect to x:
`(5x)^(3cos 2x)`
Differentiate the function with respect to x:
`(cos^(-1) x/2)/sqrt(2x+7)`, −2 < x < 2
Find `dy/dx`, if y = 12 (1 – cos t), x = 10 (t – sin t), `-pi/2 < t < pi/2`.
If f(x) = |x|3, show that f"(x) exists for all real x and find it.
Discuss the continuity and differentiability of the
If f(x) = x + 1, find `d/dx (fof) (x)`
Let f(x)= |cosx|. Then, ______.
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 ______.
cos |x| is differentiable everywhere.
`sin sqrt(x) + cos^2 sqrt(x)`
(sin x)cosx
(x + 1)2(x + 2)3(x + 3)4
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 ____________.
`d/(dx)[sin^-1(xsqrt(1 - x) - sqrt(x)sqrt(1 - x^2))]` is equal to
Let c, k ∈ R. If f(x) = (c + 1)x2 + (1 – c2)x + 2k and f(x + y) = f(x) + f(y) – xy, for all x, y ∈ R, then the value of |2(f(1) + f(2) + f(3) + ... + f(20))| is equal to ______.
A particle is moving on a line, where its position S in meters is a function of time t in seconds given by S = t3 + at2 + bt + c where a, b, c are constant. It is known that at t = 1 seconds, the position of the particle is given by S = 7 m. Velocity is 7 m/s and acceleration is 12 m/s2. The values of a, b, c are ______.
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 ______.
Let S = {t ∈ R : f(x) = |x – π| (e|x| – 1)sin |x| is not differentiable at t}. Then the set S is equal to ______.
The function f(x) = x | x |, x ∈ R is differentiable ______.
The set of all points where the function f(x) = x + |x| is differentiable, is ______.
What is \[\frac{d}{dx}(x^n)\]?
When is a function differentiable on an open interval \[(a,b)\]?
Which conclusion establishes that \[f\] is continuous at \[x=c\]?
Which statement correctly describes the converse of “differentiability implies continuity”?
For \[f(x)=|x|\], what is the right-hand derivative at \[x=0\]?
When does a derivative exist?
What does differentiability at a point mean?
