हिंदी

If x7⋅y9=(x+y)16, then show that dydx=yx - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

If `x^7 * y^9 = (x + y)^16`, then show that `dy/dx = y/x`

Find `dy/dx` if x7 . y9 = (x + y)16.

योग
Advertisements

उत्तर

x7 . y9 = (x + y)16

Taking logarithm of both sides, we get

log x7 . y9 = log (x + y)16

∴ log x7 + log y9 = 16 log (x + y)

∴ 7 log x + 9 log y = 16 log (x + y)

Differentiating both sides w.r.t. x, we get

`7(1/x) + 9(1/y) dy/dx = 16(1/(x + y)) d/dx (x + y)`

∴ `7/x + 9/y dy/dx = 16/(x + y) (1 + dy/dx)`

∴ `7/x + 9/y dy/dx = 16/(x + y) + 16/(x + y) dy/dx`

∴ `9/y dy/dx - 16/(x + y) dy/dx = 16/(x + y) - 7/x`

∴ `(9/y - 16/(x + y)) dy/dx = 16/(x + y) - 7/x`

∴ `[(9x + 9y - 16y)/(y(x + y))] dy/dx = (16x - 7x - 7y)/(x(x + y))`

∴ `[(9x - 7y)/(y(x + y))] dy/dx = (9x - 7y)/(x(x + y))`

∴ `dy/dx = (9x - 7y)/(x(x + y)) xx (y(x + y))/(9x - 7y)`

∴ `dy/dx = y/x`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Differentiation - MISCELLANEOUS EXERCISE - 3 [पृष्ठ १००]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 3 Differentiation
MISCELLANEOUS EXERCISE - 3 | Q IV] 13) | पृष्ठ १००

वीडियो ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्न

If y=eax ,show that  `xdy/dx=ylogy`


Find `bb(dy/dx)` in the following:

`y = sin^(-1)((2x)/(1+x^2))`


Find `(dy)/(dx)` if `y = sin^-1(sqrt(1-x^2))`


Find `dy/dx` if : x = 2 cos t + cos 2t, y = 2 sin t – sin 2t at t = `pi/(4)`


DIfferentiate x sin x w.r.t. tan x.


Differentiate `cos^-1((1 - x^2)/(1 + x^2)) w.r.t. tan^-1 x.`


Find `(d^2y)/(dx^2)` of the following : x = a(θ – sin θ), y = a(1 – cos θ)


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 2y = `sqrt(x + 1) + sqrt(x - 1)`, show that 4(x2 – 1)y2 + 4xy1 – y = 0.


If x2 + 6xy + y2 = 10, show that `(d^2y)/(dx^2) = (80)/(3x + y)^3`.


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:

`(1)/x`


Solve the following : 

f(x) = –x, for – 2 ≤ x < 0
= 2x, for 0 ≤ x < 2
= `(18 - x)/(4)`, for 2 < x ≤ 7
g(x) = 6 – 3x, for 0 ≤ x < 2
= `(2x - 4)/(3)`, for 2 < x ≤ 7
Let u (x) = f[g(x)], v(x) = g[f(x)] and w(x) = g[g(x)]. Find each derivative at x = 1, if it exists i.e. find u'(1), v' (1) and w'(1). If it doesn't exist, then explain why?


If `sqrt(y + x) + sqrt(y - x)` = c, show that `"dy"/"dx" = y/x - sqrt(y^2/x^2 - 1)`.


If `xsqrt(1 - y^2) + ysqrt(1 - x^2)` = 1, then show that `"dy"/"dx" = -sqrt((1 - y^2)/(1 - x^2)`.


If log y = log (sin x) – x2, show that `(d^2y)/(dx^2) + 4x "dy"/"dx" + (4x^2 + 3)y` = 0.


Choose the correct alternative.

If x = `("e"^"t" + "e"^-"t")/2, "y" = ("e"^"t" - "e"^-"t")/2`  then `"dy"/"dx"` = ? 


State whether the following is True or False:

The derivative of `"x"^"m"*"y"^"n" = ("x + y")^("m + n")` is `"x"/"y"`


Find `"dy"/"dx"` if x = `"e"^"3t",  "y" = "e"^(sqrt"t")`.


If x = sin θ, y = tan θ, then find `("d"y)/("d"x)`.


`(dy)/(dx)` of `2x + 3y = sin x` is:-


Find `(dy)/(dx)`, if `y = sin^-1 ((2x)/(1 + x^2))`


If y = y(x) is an implicit function of x such that loge(x + y) = 4xy, then `(d^2y)/(dx^2)` at x = 0 is equal to ______.


If log(x + y) = log(xy) + a then show that, `dy/dx = (-y^2)/x^2`


If log(x + y) = log(xy) + a, then show that `dy/dx = (-y^2)/x^2`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×