Advertisements
Advertisements
प्रश्न
Differentiate the following w.r.t. x : `cos^-1((sqrt(3)cosx - sinx)/(2))`
उत्तर
Let y = `cos^-1((sqrt(3)cosx - sinx)/(2))`
= `cos^-1[(cosx)((sqrt3)/2) - (sinx)(1/2)]`
= `cos^-1(cosx cos pi/6 - sinx sin pi/6) ...[∵ cos pi/6 = sqrt(3)/2, sin pi/6 = (1)/(2)]`
= `cos^-1[cos(x + pi/6)]`
= `x + pi/(6)`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"(x + pi/6)`
= `"d"/"dx"(x) + "d"/"dx"(pi/6)`
= 1 + 0
= 1.
APPEARS IN
संबंधित प्रश्न
Differentiate the following w.r.t.x: `sqrt(x^2 + 4x - 7)`
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:
tan[cos(sinx)]
Differentiate the following w.r.t.x: `log_(e^2) (log x)`
Differentiate the following w.r.t.x:
`(x^3 - 5)^5/(x^3 + 3)^3`
Differentiate the following w.r.t.x:
`sqrt(cosx) + sqrt(cossqrt(x)`
Differentiate the following w.r.t.x: `(e^sqrt(x) + 1)/(e^sqrt(x) - 1)`
Differentiate the following w.r.t.x: log[tan3x.sin4x.(x2 + 7)7]
Differentiate the following w.r.t. x :
cos3[cos–1(x3)]
Differentiate the following w.r.t. x : `cot^-1[cot(e^(x^2))]`
Differentiate the following w.r.t. x : `"cosec"^-1[1/cos(5^x)]`
Differentiate the following w.r.t. x : `"cosec"^-1((1)/(4cos^3 2x - 3cos2x))`
Differentiate the following w.r.t. x : `tan^-1[(1 + cos(x/3))/(sin(x/3))]`
Differentiate the following w.r.t. x : `cot^-1((sin3x)/(1 + cos3x))`
Differentiate the following w.r.t. x : `cos^-1((3cos3x - 4sin3x)/5)`
Differentiate the following w.r.t. x : `sin^-1((1 - x^2)/(1 + x^2))`
Differentiate the following w.r.t. x : `sin^-1(2xsqrt(1 - x^2))`
Differentiate the following w.r.t. x :
`sin^-1(4^(x + 1/2)/(1 + 2^(4x)))`
Differentiate the following w.r.t. x : `tan^-1((8x)/(1 - 15x^2))`
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 : `((x^2 + 2x + 2)^(3/2))/((sqrt(x) + 3)^3(cosx)^x`
Differentiate the following w.r.t. x:
`x^(x^x) + e^(x^x)`
Solve the following :
The values of f(x), g(x), f'(x) and g'(x) are given in the following table :
| x | f(x) | g(x) | f'(x) | fg'(x) |
| – 1 | 3 | 2 | – 3 | 4 |
| 2 | 2 | – 1 | – 5 | – 4 |
Match the following :
| A Group – Function | B Group – Derivative |
| (A)`"d"/"dx"[f(g(x))]"at" x = -1` | 1. – 16 |
| (B)`"d"/"dx"[g(f(x) - 1)]"at" x = -1` | 2. 20 |
| (C)`"d"/"dx"[f(f(x) - 3)]"at" x = 2` | 3. – 20 |
| (D)`"d"/"dx"[g(g(x))]"at"x = 2` | 5. 12 |
If y = sin−1 (2x), find `("d"y)/(""d"x)`
If y = `1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + .....,` then `(d^2y)/(dx^2)` = ______
If y = `(3x^2 - 4x + 7.5)^4, "then" dy/dx` is ______
The weight W of a certain stock of fish is given by W = nw, where n is the size of stock and w is the average weight of a fish. If n and w change with time t as n = 2t2 + 3 and w = t2 - t + 2, then the rate of change of W with respect to t at t = 1 is ______
If f(x) = `(3x + 1)/(5x - 4)` and t = `(5 + 3x)/(x - 4)`, then f(t) is ______
If y = cosec x0, then `"dy"/"dx"` = ______.
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 ______
Solve `x + y (dy)/(dx) = sec(x^2 + y^2)`
The value of `d/(dx)[tan^-1((a - x)/(1 + ax))]` is ______.
If x = eθ, (sin θ – cos θ), y = eθ (sin θ + cos θ) then `dy/dx` at θ = `π/4` is ______.
Let f(x) be a polynomial function of the second degree. If f(1) = f(–1) and a1, a2, a3 are in AP, then f’(a1), f’(a2), f’(a3) are in ______.
If y = log (sec x + tan x), find `dy/dx`.
