मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

If y = log (log 2x), show that xy2 + y1 (1 + xy1) = 0. - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

If y = log (log 2x), show that xy2 + y1 (1 + xy1) = 0.

बेरीज

उत्तर

y = log (log 2x)

∴ `"dy"/"dx" = "d"/"dx"[log(log2x)]`

= `(1)/"log2x"."d"/"dx"(log2x)`

= `(1)/"log2x" xx (1)/(2x)."d"/"dx"(2x)`

= `(1)/"log2x" xx (1)/(2x) xx 2`

∴ `"dy"/"dx" = (1)/(xlog2x)`

∴ `(log2x)."dy"/"dx" = (1)/x`                 ...(1)
Differentiating both sides w.r.t. x, we get

`(log2x)."d"/"dx"(dx/dy) + "dy"/"dx"."d"/"dx"(log2x) = "d"/"dx"(1/x)`

∴ `(log2x).(d^2y)/(dx^2) + "dy"/"dx".(1)/(2x)."d"/"dx"(2x) = -(1)/x^2`

∴ `(log2x).(d^2y)/(dx^2) + "dy"/"dx".(1)/(2x) xx 2 = -(1)/x^2`

∴ `(log2x).(d^2y)/(dx^2) + (1)/x."dy"/"dx" = (1)/x.(1)/x`

∴ `(log2x).(d^2y)/(dx^2) + [(log2x)."dy"/"dx"]"dy"/"dx" = -(1)/x[(log2x)."dy"/"dx"]`     ...[By (1)]

Dividing throughout by log 2x, we get

`(d^2y)/(dx^2) + (dy/dx)^2 = -(1)/x"dy"/"dx"`

∴ `x(d^2y)/(dx^2) + x(dy/dx)^2 = -"dy"/"dx"`

∴ `x(d^2y)/(dx^2) + "dy"/"dx" + x(dy/dx)^2` = 0

∴ `x(d^2y)/(dx^2) + "dy"/"dx" (1 + xdy/dx)` = 0

∴ xy2 + y1 (1 + xy1) = 0.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Differentiation - Exercise 1.5 [पृष्ठ ६०]

APPEARS IN

संबंधित प्रश्‍न

 

if xx+xy+yx=ab, then find `dy/dx`.


Differentiate the function with respect to x. 

cos x . cos 2x . cos 3x


Differentiate the function with respect to x.

`sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))`


Differentiate the function with respect to x.

`x^x - 2^(sin x)`


Differentiate the function with respect to x.

`(x cos x)^x + (x sin x)^(1/x)`


Find `dy/dx`for the function given in the question:

xy + yx = 1


Differentiate w.r.t. x the function:

xx + xa + ax + aa, for some fixed a > 0 and x > 0


If `y = sin^-1 x + cos^-1 x , "find"  dy/dx`


Find `(dy)/(dx) , if y = sin ^(-1) [2^(x +1 )/(1+4^x)]`


Evaluate 
`int  1/(16 - 9x^2) dx`


Find `dy/dx` if y = x+ 5x


If `(sin "x")^"y" = "x" + "y", "find" (d"y")/(d"x")`


If log (x + y) = log(xy) + p, where p is a constant, then prove that `"dy"/"dx" = (-y^2)/(x^2)`.


If `log_10((x^3 - y^3)/(x^3 + y^3)) = 2, "show that" "dy"/"dx" = -(99x^2)/(101y^2)`


If `log_5((x^4 + y^4)/(x^4 - y^4)) = 2, "show that""dy"/"dx" = (12x^3)/(13y^3)`.


If xy = ex–y, then show that `"dy"/"dx" = logx/(1 + logx)^2`.


If y = `x^(x^(x^(.^(.^.∞))`, then show that `"dy"/"dx" = y^2/(x(1 - logy).`.


If x = `asqrt(secθ - tanθ), y = asqrt(secθ + tanθ), "then show that" "dy"/"dx" = -y/x`.


If x = a cos3t, y = a sin3t, show that `"dy"/"dx" = -(y/x)^(1/3)`.


Find the second order derivatives of the following : log(logx)


If y = `log(x + sqrt(x^2 + a^2))^m`, show that `(x^2 + a^2)(d^2y)/(dx^2) + x "d"/"dx"` = 0.


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


If f(x) = logx (log x) then f'(e) is ______


If y = `log[4^(2x)((x^2 + 5)/sqrt(2x^3 - 4))^(3/2)]`, find `("d"y)/("d"x)`


The rate at which the metal cools in moving air is proportional to the difference of temperatures between the metal and air. If the air temperature is 290 K and the metal temperature drops from 370 K to 330 K in 1 O min, then the time required to drop the temperature upto 295 K.


lf y = `2^(x^(2^(x^(...∞))))`, then x(1 - y logx logy)`dy/dx` = ______  


If y = tan-1 `((1 - cos 3x)/(sin 3x))`, then `"dy"/"dx"` = ______.


`"d"/"dx" [(cos x)^(log x)]` = ______.


Derivative of `log_6`x with respect 6x to is ______


`lim_("x" -> 0)(1 - "cos x")/"x"^2` is equal to ____________.


`lim_("x" -> -2) sqrt ("x"^2 + 5 - 3)/("x" + 2)` is equal to ____________.


If y `= "e"^(3"x" + 7), "then the value" |("dy")/("dx")|_("x" = 0)` is ____________.


If `f(x) = log [e^x ((3 - x)/(3 + x))^(1/3)]`,  then `f^'(1)` is equal to


Given f(x) = `log((1 + x)/(1 - x))` and g(x) = `(3x + x^3)/(1 + 3x^2)`, then fog(x) equals


If `log_10 ((x^3 - y^3)/(x^3 + y^3))` = 2 then `dy/dx` = ______.


If `log_10 ((x^2 - y^2)/(x^2 + y^2))` = 2, then `dy/dx` is equal to ______.


Find the derivative of `y = log x + 1/x` with respect to x.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×