हिंदी

If x = t2, y = t3, then dydxd2ydx2 is ______. - Mathematics

Advertisements
Advertisements

प्रश्न

If x = t2, y = t3, then `("d"^2"y")/("dx"^2)` is ______.

विकल्प

  • `3/2`

  • `3/(4"t")`

  • `3/(2"t")`

  • `3/4`

MCQ
रिक्त स्थान भरें
Advertisements

उत्तर

If x = t2, y = t3, then `("d"^2"y")/("dx"^2)` is `3/(4"t")`.

Explanation:

Given that x = t2 and y = t3 

Differentiating both the parametric functions w.r.t. t

`"dx"/"dt"` = 2t and  `"dy"/"dt"` = 3t2

∴ `"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`

= `(3"t"^2)/(2"t")`

= `3/2 "t"`

⇒ `"dy"/"dx" = 3/2 "t"`

Now differentiating again w.r.t. x

`"d"/"dx"("dy"/"dx") = 3/2 * "dt"/"dx"`

⇒ `("d"^2"y")/("dx"^2) = 3/2 * 1/(2"t")`

= `3/(4"t")`.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Continuity And Differentiability - Exercise [पृष्ठ ११५]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 5 Continuity And Differentiability
Exercise | Q 94 | पृष्ठ ११५

संबंधित प्रश्न

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


If x = f(t), y = g(t) are differentiable functions of parammeter ‘ t ’ then prove that y is a differentiable function of 'x' and  hence, find dy/dx if x=a cost, y=a sint


If `x=a(t-1/t),y=a(t+1/t)`, then show that `dy/dx=x/y`


If `ax^2+2hxy+by^2=0` , show that `(d^2y)/(dx^2)=0`


If y =1 − cos θ, x = 1 − sin θ, then `dy/dx  "at"  θ =pi/4` is ______


If x=α sin 2t (1 + cos 2t) and y=β cos 2t (1cos 2t), show that `dy/dx=β/αtan t`


If x = a sin 2t (1 + cos2t) and y = b cos 2t (1 – cos 2t), find the values of  `dy/dx `at t = `pi/4`


If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t) then find `dy/dx `

 


If x = cos t (3 – 2 cos2 t) and y = sin t (3 – 2 sin2 t), find the value of dx/dy at t =4/π.


Derivatives of  tan3θ with respect to sec3θ at θ=π/3 is

(A)` 3/2`

(B) `sqrt3/2`

(C) `1/2`

(D) `-sqrt3/2`


If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = sin t, y = cos 2t


If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = 4t, y = `4/y`


If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = cos θ – cos 2θ, y = sin θ – sin 2θ


If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = a (θ – sin θ), y = a (1 + cos θ)


If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

`x = (sin^3t)/sqrt(cos 2t), y = (cos^3t)/sqrt(cos 2t)`


If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = a sec θ, y = b tan θ


If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)


If x = `sqrt(a^(sin^(-1)t))`, y = `sqrt(a^(cos^(-1)t))` show that `dy/dx = - y/x`.


If `x = acos^3t`, `y = asin^3 t`,

Show that `(dy)/(dx) =- (y/x)^(1/3)`


If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find `dy/dx` when `theta = pi/3`


IF `y = e^(sin-1x)   and  z =e^(-cos-1x),` prove that `dy/dz = e^x//2`


The cost C of producing x articles is given as C = x3-16x2 + 47x.  For what values of x, with the average cost is decreasing'?  


Evaluate : `int  (sec^2 x)/(tan^2 x + 4)` dx


x = `"t" + 1/"t"`, y = `"t" - 1/"t"`


x = `"e"^theta (theta + 1/theta)`, y= `"e"^-theta (theta - 1/theta)`


x = 3cosθ – 2cos3θ, y = 3sinθ – 2sin3θ


sin x = `(2"t")/(1 + "t"^2)`, tan y = `(2"t")/(1 - "t"^2)`


x = `(1 + log "t")/"t"^2`, y = `(3 + 2 log "t")/"t"`


If x = 3sint – sin 3t, y = 3cost – cos 3t, find `"dy"/"dx"` at t = `pi/3`


Differentiate `tan^-1 ((sqrt(1 + x^2) - 1)/x)` w.r.t. tan–1x, when x ≠ 0


If `"x = a sin"  theta  "and  y = b cos"  theta, "then"  ("d"^2 "y")/"dx"^2` is equal to ____________.


If y `= "Ae"^(5"x") + "Be"^(-5"x") "x"  "then"  ("d"^2 "y")/"dx"^2` is equal to ____________.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×