Advertisements
Advertisements
प्रश्न
Find the rate of change of demand (x) of a commodity with respect to its price (y) if y = 25 + 30x – x2.
उत्तर
y = 25 + 30x – x2.
Differentiating both sides w.r.t. x, we get
`"dy"/"dx" = "d"/"dx" (25 + 30"x" - "x"^2) = 0 + 30 - 2"x"`
∴ `"dy"/"dx" = 30 - 2"x"`
Now, by the derivative of an inverse function, the rate of change of demand (x) w.r.t. price(y) is
`"dx"/"dy" = 1/(("dy"/"dx"))`, where `"dy"/"dx" ne 0`.
i.e. `"dx"/"dy" = 1/(30 - 2"x")`
APPEARS IN
संबंधित प्रश्न
If y = eax. cos bx, then prove that
`(d^2y)/(dx^2) - 2ady/dx + (a^2 + b^2)y` = 0
Solve : `"dy"/"dx" = 1 - "xy" + "y" - "x"`
Find `"dy"/"dx"` if `xsqrt(x) + ysqrt(y) = asqrt(a)`
Find `"dy"/"dx"` if xey + yex = 1
Find `"dy"/"dx"` if ex+y = cos(x – y)
Find `"dy"/"dx"` if cos (xy) = x + y
`d/dx(10^x) = x*10^(x - 1)`
Find `"dy"/"dx"`, if y = xx.
If y = cos−1 [sin (4x)], find `("d"y)/("d"x)`
Choose the correct alternative:
If y = `x^(sqrt(x))`, then `("d"y)/("d"x)` = ?
y = (6x4 – 5x3 + 2x + 3)6, find `("d"y)/("d"x)`
Solution: Given,
y = (6x4 – 5x3 + 2x + 3)6
Let u = `[6x^4 - 5x^3 + square + 3]`
∴ y = `"u"^square`
∴ `("d"y)/"du"` = 6u6–1
∴ `("d"y)/"du"` = 6( )5
and `"du"/("d"x) = 24x^3 - 15(square) + 2`
By chain rule,
`("d"y)/("d"x) = ("d"y)/square xx square/("d"x)`
∴ `("d"y)/("d"x) = 6(6x^4 - 5x^3 + 2x + 3)^square xx (24x^3 - 15x^2 + square)`
If u = x2 + y2 and x = s + 3t, y = 2s - t, then `(d^2u)/(ds^2)` = ______
If y = `x/"e"^(1 + x)`, then `("d"y)/("d"x)` = ______.
If f(x) = |cos x|, find f'`((3pi)/4)`
If f(x) = `{{:(x^3 + 1",", x < 0),(x^2 + 1",", x ≥ 0):}`, g(x) = `{{:((x - 1)^(1//3)",", x < 1),((x - 1)^(1//2)",", x ≥ 1):}`, then (gof) (x) is equal to ______.
If y = 2x2 + a2 + 22 then `dy/dx` = ______.
If y = `log((x + sqrt(x^2 + a^2))/(sqrt(x^2 + a^2) - x))`, find `dy/dx`.
Find `dy/dx` if, y = `e^(5x^2-2x+4)`
If `y = (x + sqrt(a^2 + x^2))^m`, prove that `(a^2 + x^2)(d^2y)/(dx^2) + xdy/dx - m^2y = 0`
