मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

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.

बेरीज
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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Differentiation - MISCELLANEOUS EXERCISE - 3 [पृष्ठ १००]

APPEARS IN

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

व्हिडिओ ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्‍न

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

sin2 y + cos xy = k


Show that the derivative of the function f given by 

\[f\left( x \right) = 2 x^3 - 9 x^2 + 12x + 9\], at x = 1 and x = 2 are equal.

If  \[f\left( x \right) = x^3 + 7 x^2 + 8x - 9\] 

, find f'(4).


Is |sin x| differentiable? What about cos |x|?


Let \[f\left( x \right)\begin{cases}a x^2 + 1, & x > 1 \\ x + 1/2, & x \leq 1\end{cases}\] . Then, f (x) is derivable at x = 1, if 


Find `(dy)/(dx) , "If"   x^3 + y^2 + xy = 10`


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


Discuss extreme values of the function f(x) = x.logx


Find `"dy"/"dx"` if : x = a cos3θ, y = a sin3θ at θ = `pi/(3)`


Differentiate `tan^-1((cosx)/(1 + sinx)) w.r.t. sec^-1 x.`


Differentiate xx w.r.t. xsix.


If y = `e^(mtan^-1x)`, show that `(1 + x^2)(d^2y)/(dx^2) + (2x - m)"dy"/"dx"` = 0.


If `sec^-1((7x^3 - 5y^3)/(7^3 + 5y^3)) = "m", "show"  (d^2y)/(dx^2)` = 0.


Find the nth derivative of the following : (ax + b)m 


Find the nth derivative of the following : apx+q 


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 = sec (tan –1x), then `"dy"/"dx"` at x = 1, is equal to


Choose the correct option from the given alternatives :

If f(x) = `sin^-1((4^(x + 1/2))/(1 + 2^(4x)))`, which of the following is not the derivative of f(x)?


Choose the correct option from the given alternatives :

If `xsqrt(y + 1) + ysqrt(x + 1) = 0 and x ≠ y, "then" "dy"/"dx"` = ........


Differentiate the following w.r.t. x : `sin[2tan^-1(sqrt((1 - x)/(1 + x)))]`


Differentiate the following w.r.t. x:

`tan^-1(x/(1 + 6x^2)) + cot^-1((1 - 10x^2)/(7x))`


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


If x sin (a + y) + sin a . cos (a + y) = 0, then show that `"dy"/"dx" = (sin^2(a + y))/(sina)`.


Differentiate log `[(sqrt(1 + x^2) + x)/(sqrt(1 + x^2 - x)]]` w.r.t. cos (log x).


Find `"dy"/"dx"` if, `"x"^"y" = "e"^("x - y")`


Differentiate w.r.t x (over no. 24 and 25) `e^x/sin x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×