Advertisements
Advertisements
प्रश्न
Choose the correct alternative.
If `"x"^4."y"^5 = ("x + y")^("m + 1")` then `"dy"/"dx" = "y"/"x"` then m = ?
पर्याय
8
4
5
20
Advertisements
उत्तर
8
Explanation:
x4. y5 = (x + y)m + 1 ...(i)
∴ `"d"/"dx" ("x"^4. "y"^5) = "d"/"dx" ("x" + "y")^("m" + 1)`
∴ `"x"^4 "d"/"dx" "y"^5 + "y"^5 "d"/"dx" "x"^4 = ("m" + 1)("x" + "y")^("m" + 1 − 1) . "d"/"dx" ("x" + "y")`
∴ `"x"^4 . 5"y"^4 "d"/"dx" "y"+ "y"^5 4"x"^3 "d"/"dx" "x" = ("m" + 1)("x" + "y")^"m" ["d"/"dx" "x" + "d"/"dx" "y"]`
∴ `5"x"^4"y"^4 "dy"/"dx" + 4"x"^3 "y"^5 . 1 = ("m" + 1)("x" + "y")^"m" [1 + "dy"/"dx"]`
∴ `5"x"^4"y"^4 "dy"/"dx" + 4"x"^3 "y"^5 = ("m" + 1)("x" + "y")^"m" [1 + "dy"/"dx"]`
Put `"dy"/"dx" = "y"/"x"`
∴ `5"x"^((cancel4)3)"y"^4 . "y"/cancel"x" + 4"x"^3 "y"^5 = ("m" + 1)("x" + "y")^"m" [1 + "y"/"x"]`
∴ `5"x"^3"y"^4 . "y" + 4"x"^3 "y"^5 = ("m" + 1)("x" + "y")^"m" [("x" + "y")/"x"]`
∴ `5"x"^3"y"^5 + 4"x"^3 "y"^5 = ("m" + 1)("x" + "y")^"m" [("x" + "y")/"x"]`
∴ `9"x"^3"y"^5 = ("m" + 1)/"x" [("x" + "y")^("m" + 1)]`
∴ `9"x"^3"y"^5 = ("m" + 1)/cancel"x" "x"^((cancel4)3)."y"^5`
∴ `9cancel("x"^3"y"^5) = ("m" + 1) cancel("x"^3"y"^5)`
∴ 9 = m + 1
∴ m = 9 - 1
∴ m = 8
APPEARS IN
संबंधित प्रश्न
Find `bb(dy/dx)` in the following:
2x + 3y = sin y
Write the derivative of f (x) = |x|3 at x = 0.
Find `(dy)/(dx) , "If" x^3 + y^2 + xy = 10`
Differentiate tan-1 (cot 2x) w.r.t.x.
Find `"dy"/"dx"` if x = at2, y = 2at.
Find `"dy"/"dx"` if x = a cot θ, y = b cosec θ
Find `"dy"/"dx"`, if : x = sinθ, y = tanθ
Find `"dy"/"dx"`, if : x = `(t + 1/t)^a, y = a^(t+1/t)`, where a > 0, a ≠ 1, t ≠ 0.
Find `"dy"/"dx"`, if : `x = cos^-1(4t^3 - 3t), y = tan^-1(sqrt(1 - t^2)/t)`.
Find `"dy"/"dx"` if : x = t + 2sin (πt), y = 3t – cos (πt) at t = `(1)/(2)`
Differentiate `tan^-1((x)/(sqrt(1 - x^2))) w.r.t. sec^-1((1)/(2x^2 - 1))`.
Differentiate `tan^-1((cosx)/(1 + sinx)) w.r.t. sec^-1 x.`
Find `(d^2y)/(dx^2)` of the following : x = a cos θ, y = b sin θ at θ = `π/4`.
If x = at2 and y = 2at, then show that `xy(d^2y)/(dx^2) + a` = 0.
If x = cos t, y = emt, show that `(1 - x^2)(d^2y)/(dx^2) - x"dy"/"dx" - m^2y` = 0.
If y = x + tan x, show that `cos^2x.(d^2y)/(dx^2) - 2y + 2x` = 0.
If y = eax.sin(bx), show that y2 – 2ay1 + (a2 + b2)y = 0.
If x = a sin t – b cos t, y = a cos t + b sin t, show that `(d^2y)/(dx^2) = -(x^2 + y^2)/(y^3)`.
Find the nth derivative of the following : sin (ax + b)
Find the nth derivative of the following:
y = e8x . cos (6x + 7)
Choose the correct option from the given alternatives :
If y = sin (2sin–1 x), then dx = ........
If x = `e^(x/y)`, then show that `dy/dx = (x - y)/(xlogx)`
Differentiate `tan^-1((sqrt(1 + x^2) - 1)/x)` w.r.t. `cos^-1(sqrt((1 + sqrt(1 + x^2))/(2sqrt(1 + x^2))))`
Find `"dy"/"dx"` if, x3 + y3 + 4x3y = 0
Find `"dy"/"dx"` if, `"x"^"y" = "e"^("x - y")`
If `x^7 * y^9 = (x + y)^16`, then show that `dy/dx = y/x`
If y = `sqrt(tansqrt(x)`, find `("d"y)/("d"x)`.
If `tan ((x + y)/(x - y))` = k, then `dy/dx` is equal to ______.
