Advertisements
Advertisements
प्रश्न
If x = A cos 4t + B sin 4t, then `(d^2x)/(dt^2)` is equal to ______.
विकल्प
x
– x
16x
– 16x
Advertisements
उत्तर
If x = A cos 4t + B sin 4t, then `(d^2x)/(dt^2)` is equal to – 16x.
Explanation:
x = A cos 4t + B sin 4t
`dx/dt` = – A 4 sin 4t + 4B cos 4t
`(d^2x)/(dt^2)` = – 16A cos 4t – 16B sin 4t
= – 16[A cos 4t + B sin 4t]
= – 16x.
APPEARS IN
संबंधित प्रश्न
If x = a sin t and `y = a (cost+logtan(t/2))` ,find `((d^2y)/(dx^2))`
Find the second order derivative of the function.
x20
Find the second order derivative of the function.
x . cos x
Find the second order derivative of the function.
ex sin 5x
Find the second order derivative of the function.
e6x cos 3x
If y = 5 cos x – 3 sin x, prove that `(d^2y)/(dx^2) + y = 0`.
If y = (tan–1 x)2, show that (x2 + 1)2 y2 + 2x (x2 + 1) y1 = 2
Find `("d"^2"y")/"dx"^2`, if y = `"x"^5`
Find `("d"^2"y")/"dx"^2`, if y = `"e"^"x"`
Find `("d"^2"y")/"dx"^2`, if y = 2at, x = at2
If ax2 + 2hxy + by2 = 0, then show that `("d"^2"y")/"dx"^2` = 0
`sin xy + x/y` = x2 – y
sec(x + y) = xy
Read the following passage and answer the questions given below:
|
The relation between the height of the plant ('y' in cm) with respect to its exposure to the sunlight is governed by the following equation y = `4x - 1/2 x^2`, where 'x' is the number of days exposed to the sunlight, for x ≤ 3.
|
- Find the rate of growth of the plant with respect to the number of days exposed to the sunlight.
- Does the rate of growth of the plant increase or decrease in the first three days? What will be the height of the plant after 2 days?
Find `(d^2y)/dx^2` if, `y = e^((2x + 1))`
Find `(d^2y)/dx^2` if, y = `e^((2x + 1))`
Find `(d^2y)/dx^2` if, `y = e^((2x + 1))`
Find `(d^2y)/dx^2` if, y = `e^(2x +1)`
Find `(d^2y)/(dx^2) "if", y = e^((2x + 1))`
When is the second order derivative of \[y\] with respect to \[x\] defined?
If \[y=\mathrm{A}\sin x+\mathrm{B}\cos x\], what is \[\frac{dy}{dx}\]?
For \[y=\mathrm{A}\sin x+\mathrm{B}\cos x\], what is \[\frac{d^2y}{dx^2}\]?
If \[y=\sin^{-1}x\], what is \[\frac{dy}{dx}\]?
Which equation is equivalent to \[\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\] for \[y=\sin^{-1}x\]?
After differentiating \[\sqrt{1-x^2}\cdot\frac{dy}{dx}=1\], which equation results?
For \[y=\sin^{-1}x\], which relation uses \[y_{1}\] for the first derivative?

