हिंदी

If xm . yn = (x + y)m+n, prove that dydxd2ydx2 = 0

Advertisements
Advertisements

प्रश्न

If xm . yn = (x + y)m+n, prove that `("d"^2"y")/("dx"^2)` = 0

योग
Advertisements

उत्तर

Given that: `"dy"/"dx" = y/x`

Differentiating both sides w.r.t. x

`"d"/"dx"("dy"/"dx") = "d"/"dx"(y/x)`

⇒ `("d"^2y)/("dx"^2) = (x* "dy"/"dx" y*1)/x^2`

= `(x * y/x - 1)/x^2`   .....`[because "dy"/"dx" = y/x]`

= `(y - y)/x^2`

= `0/x^2`

= 0

Hence, `("d"^2y)/("dx"^2)` = 0.

Hence proved.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Continuity And Differentiability - Exercise [पृष्ठ ११३]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 5 Continuity And Differentiability
Exercise | Q 80. (ii) | पृष्ठ ११३

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Differentiate the function with respect to x.

cos (sin x)


Differentiate the function with respect to x.

sin (ax + b)


Differentiate the function with respect to x.

`sec(tan (sqrtx))`


Differentiate the function with respect to x.

`cos (sqrtx)`


Differentiate the function with respect to x:

sin3 x + cos6 x


Find `dy/dx`, if y = 12 (1 – cos t), x = 10 (t – sin t), `-pi/2 < t < pi/2`.


If f(x) = |x|3, show that f"(x) exists for all real x and find it.


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


If y = tanx + secx, prove that `("d"^2y)/("d"x^2) = cosx/(1 - sinx)^2`


COLUMN-I COLUMN-II
(A) If a function
f(x) = `{((sin3x)/x, "if"  x = 0),("k"/2",",  "if"  x = 0):}`
is continuous at x = 0, then k is equal to
(a) |x|
(B) Every continuous function is differentiable (b) True
(C) An example of a function which is continuous
everywhere but not differentiable at exactly one point
(c) 6
(D) The identity function i.e. f (x) = x ∀ ∈x R
is a continuous function
(d) False

`cos(tan sqrt(x + 1))`


`sin^-1  1/sqrt(x + 1)`


(sin x)cosx 


`cos^-1 ((sinx + cosx)/sqrt(2)), (-pi)/4 < x < pi/4`


`tan^-1 (sqrt((1 - cosx)/(1 + cosx))), - pi/4 < x < pi/4`


`tan^-1 (("a"cosx - "b"sinx)/("b"cosx - "a"sinx)), - pi/2 < x < pi/2` and `"a"/"b" tan x > -1`


`sec^-1 (1/(4x^3 - 3x)), 0 < x < 1/sqrt(2)`


`tan^-1 ((3"a"^2x - x^3)/("a"^3 - 3"a"x^2)), (-1)/sqrt(3) < x/"a" < 1/sqrt(3)`


`tan^-1 ((sqrt(1 + x^2) + sqrt(1 - x^2))/(sqrt(1 + x^2) - sqrt(1 - x^2))), -1 < x < 1, x ≠ 0`


For the curve `sqrt(x) + sqrt(y)` = 1, `"dy"/"dx"` at `(1/4, 1/4)` is ______.


The differential coefficient of `"tan"^-1 ((sqrt(1 + "x") - sqrt (1 - "x"))/(sqrt (1+ "x") + sqrt (1 - "x")))` is ____________.


If `"f"("x") = ("sin" ("e"^("x"-2) - 1))/("log" ("x" - 1)), "x" ne 2 and "f" ("x") = "k"` for x = 2, then value of k for which f is continuous is ____________.


If `y = (x + sqrt(1 + x^2))^n`, then `(1 + x^2) (d^2y)/(dx^2) + x (dy)/(dx)` is


Let c, k ∈ R. If f(x) = (c + 1)x2 + (1 – c2)x + 2k and f(x + y) = f(x) + f(y) – xy, for all x, y ∈ R, then the value of |2(f(1) + f(2) + f(3) + ... + f(20))| is equal to ______.


A particle is moving on a line, where its position S in meters is a function of time t in seconds given by S = t3 + at2 + bt + c where a, b, c are constant. It is known that at t = 1 seconds, the position of the particle is given by S = 7 m. Velocity is 7 m/s and acceleration is 12 m/s2. The values of a, b, c are ______.


The function f(x) = x | x |, x ∈ R is differentiable ______.


The set of all points where the function f(x) = x + |x| is differentiable, is ______.


Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.


If \[u\] and \[v\] are differentiable functions, which formula is correct?


What is \[\frac{d}{dx}(\sin x)\]?


What is \[\frac{d}{dx}(\tan x)\]?


A function is differentiable at \[x=c\] when which condition holds?


What is the practical condition for differentiability at a point?


When is a function differentiable on an open interval \[(a,b)\]?


If \[f\] is differentiable at \[c\], which limit equals \[f'(c)\]?


Which conclusion establishes that \[f\] is continuous at \[x=c\]?


What does differentiability at a point mean?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×