Advertisements
Advertisements
प्रश्न
Find the second order derivative of the function.
e6x cos 3x
उत्तर
Let, y = e6x cos 3x
Differentiating both sides with respect to x,
`dy/dx = e^(6x) d/dx cos 3 x + cos 3 x d/dx e^(6x)`
`= e^(6x) (- sin 3 x) d/dx (3x) + cos 3 x * e^(6x) d/dx (6x)`
`= - 3 e^(6x) sin 3 x + 6 e^(6x) cos 3 x`
`= e^(6x) (6 cos 3 x - 3 sin 3 x)`
Differentiating both sides again with respect to x,
`(d^2 y)/dx^2 = e^(6x) d/dx (6 cos 3 x - 3 sin 3 x) + (6 cos 3 x - 3 sin 3 x) d/dx e^(6x)`
`= e^(6x) [6 (- sin 3x) d/dx (3x) - 3 cos 3x d/dx (3x)] + [6 cos 3x - 3 sin 3x]e^(6x) d/dx (6x)`
`= e^(6x) [-6 sin 3x * 3 - 3 cos 3x . 3] + [6 cos 3x - 3 sin 3x] xx e^(6x) * 6`
`= e^(6x) [- 18 sin 3x - 9 cos 3 x] + e^(6x) [36 cos 3x - 18 sin 3x]`
`= e^(6x) [- 36 sin 3x + 27 cos 3x]`
`= 9 e^(6x) [3 cos 3x - 4 sin 3x]`
APPEARS IN
संबंधित प्रश्न
If x = a sin t and `y = a (cost+logtan(t/2))` ,find `((d^2y)/(dx^2))`
If x cos(a+y)= cosy then prove that `dy/dx=(cos^2(a+y)/sina)`
Hence show that `sina(d^2y)/(dx^2)+sin2(a+y)(dy)/dx=0`
Find the second order derivative of the function.
x2 + 3x + 2
Find the second order derivative of the function.
log x
Find the second order derivative of the function.
ex sin 5x
Find the second order derivative of the function.
log (log x)
Find the second order derivative of the function.
sin (log x)
If y = 5 cos x – 3 sin x, prove that `(d^2y)/(dx^2) + y = 0`
If y = cos–1 x, Find `(d^2y)/dx^2` in terms of y alone.
If y = 3 cos (log x) + 4 sin (log x), show that x2 y2 + xy1 + y = 0
If y = 500e7x + 600e–7x, show that `(d^2y)/(dx^2) = 49y`
If ey (x + 1) = 1, show that `(d^2y)/(dx^2) =((dy)/(dx))^2`
Find `("d"^2"y")/"dx"^2`, if y = `sqrt"x"`
Find `("d"^2"y")/"dx"^2`, if y = `"x"^-7`
Find `("d"^2"y")/"dx"^2`, if y = log (x).
Find `("d"^2"y")/"dx"^2`, if y = 2at, x = at2
Find `("d"^2"y")/"dx"^2`, if y = `"x"^2 * "e"^"x"`
If ax2 + 2hxy + by2 = 0, then show that `("d"^2"y")/"dx"^2` = 0
tan–1(x2 + y2) = a
If x sin (a + y) + sin a cos (a + y) = 0, prove that `"dy"/"dx" = (sin^2("a" + y))/sin"a"`
If y = tan–1x, find `("d"^2y)/("dx"^2)` in terms of y alone.
The derivative of cos–1(2x2 – 1) w.r.t. cos–1x is ______.
If y = 5 cos x – 3 sin x, then `("d"^2"y")/("dx"^2)` is equal to:
If x2 + y2 + sin y = 4, then the value of `(d^2y)/(dx^2)` at the point (–2, 0) is ______.
If x = A cos 4t + B sin 4t, then `(d^2x)/(dt^2)` is equal to ______.
If x = a cos t and y = b sin t, then find `(d^2y)/(dx^2)`.
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))`
Find `(d^2y)/(dx^2) "if", y = e^((2x + 1))`
