Advertisements
Advertisements
प्रश्न
Solve `x + y (dy)/(dx) = sec(x^2 + y^2)`
Advertisements
उत्तर
Given, `x + y (dy)/(dx) = sec(x^2 + y^2)` .....(i)
Put x2 + y2 = t
Differentiating w.r.t. x, we get
`2x + 2y (dy)/(dx) = (dt)/(dx)`
⇒ `x + y (dy)/(dx) = 1/2 (dt)/(dx)`
Put this value in (i), we get
`1/2 (dt)/(dx)` = sec t
⇒ `(dt)/sec t` = 2 dx
Integrating on both sides, we get
`int cost dt = 2 int dx`
⇒ sin t = 2x + c
⇒ sin(x2 + y2) = 2x + C ......(∵ t = x2 + y2)
is the required solution.
APPEARS IN
संबंधित प्रश्न
Differentiate the following w.r.t.x: cos(x2 + a2)
Differentiate the following w.r.t.x: `"cosec"(sqrt(cos x))`
Differentiate the following w.r.t.x: log[cos(x3 – 5)]
Differentiate the following w.r.t.x: `e^(log[(logx)^2 - logx^2]`
Differentiate the following w.r.t.x: `log[sec (e^(x^2))]`
Differentiate the following w.r.t.x: (1 + sin2 x)2 (1 + cos2 x)3
Differentiate the following w.r.t.x:
`sqrt(cosx) + sqrt(cossqrt(x)`
Differentiate the following w.r.t.x: `cot(logx/2) - log(cotx/2)`
Differentiate the following w.r.t.x:
`log(sqrt((1 + cos((5x)/2))/(1 - cos((5x)/2))))`
Differentiate the following w.r.t.x: `log[4^(2x)((x^2 + 5)/(sqrt(2x^3 - 4)))^(3/2)]`
Differentiate the following w.r.t.x:
`log[a^(cosx)/((x^2 - 3)^3 logx)]`
Differentiate the following w.r.t. x : tan–1(log x)
Differentiate the following w.r.t. x : `tan^-1(sqrt(x))`
Differentiate the following w.r.t. x :
cos3[cos–1(x3)]
Differentiate the following w.r.t. x : `sin^4[sin^-1(sqrt(x))]`
Differentiate the following w.r.t. x : `cot^-1[cot(e^(x^2))]`
Differentiate the following w.r.t. x :
`cos^-1(sqrt(1 - cos(x^2))/2)`
Differentiate the following w.r.t. x : `cos^-1((3cos3x - 4sin3x)/5)`
Differentiate the following w.r.t. x : `"cosec"^-1[(10)/(6sin(2^x) - 8cos(2^x))]`
Differentiate the following w.r.t. x :
`cos^-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 : `tan^-1((8x)/(1 - 15x^2))`
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((5 -x)/(6x^2 - 5x - 3))`
Differentiate the following w.r.t. x : `cot^-1((4 - x - 2x^2)/(3x + 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^(e^x) + (logx)^(sinx)`
Differentiate the following w.r.t. x :
(sin x)tanx + (cos x)cotx
Differentiate the following w.r.t. x : `10^(x^(x)) + x^(x(10)) + x^(10x)`
Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : `cos^-1((7x^4 + 5y^4)/(7x^4 - 5y^4)) = tan^-1a`
Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : `sin((x^3 - y^3)/(x^3 + y^3))` = a3
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 |
Differentiate y = etanx w.r. to x
If f(x) is odd and differentiable, then f′(x) is
Differentiate `sin^-1((2cosx + 3sinx)/sqrt(13))` w.r. to x
If y = `sqrt(cos x + sqrt(cos x + sqrt(cos x + ...... ∞)`, show that `("d"y)/("d"x) = (sin x)/(1 - 2y)`
If x = `sqrt("a"^(sin^-1 "t")), "y" = sqrt("a"^(cos^-1 "t")), "then" "dy"/"dx"` = ______
If `t = v^2/3`, then `(-v/2 (df)/dt)` is equal to, (where f is acceleration) ______
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 ______.
If `cos((x^2 - y^2)/(x^2 + y^2))` = log a, show that `dy/dx = y/x`
