y= x cos x ; Find (d^2y/dx^2)?

Answer 1

y= x cos x

=> dy/dx
= x * d/dx (cos x) + (cos x) * d/dx (x)
= - x sin x + cos x

d^2y/dx^2
= d/dx (dy/dx)
= d/dx ( - x sin x + cos x)
= - x * d/dx (sin x) - sin x * d/dx (x) + d/dx (cos x)
= - x cos x - sin x - sin x
= - x cos x - 2 sin x

Answer 2

Using the product rule of differentiation, we have for the first derivative
dy / dx = d[sin x cos x] / dx
= (cos x) d[sin x]/dx + (sin x) d[cos x]/dx [Product Rule]
= (cos x ) cos x + (sin x) (-sin x)
= cos^2 x - sin^2 x

The second derivative
d^2y / dx^2 = d[dy/dx] / dx = d[cos^2 x - sin^2 x] / dx
= d[cos^2 x] / dx - d[sin^2 x] / dx

Now, we will apply the famous Chain Rule. First, let
u = cos x, v = sin x
Then
d[cos^2 x] / dx = du^2 / dx
= (du^2 / du) (du / dx) [Chain Rule]
= (2u) (d cos x / dx)
= (2cos x) ( - sin x)
= -2sin x cos x

d[sin^2 x] / dx = dv^2 / dx
= (dv^2 / dv) (dv / dx) [Chain Rule]
= (2v) (d[sin x] / dx)
= (2sin x) (cos x)
= 2sin x cos x

Finally, d^2y / dx^2 = d[cos^2 x] / dx - d[sin^2 x] / dx
= -2sin x cos x - 2sin x cos x
= -4 sin x cos x

Or, we can apply the Chain Rule directly to the second derivative: d^2y / dx^2 = d[cos^2 x] / dx - d[sin^2 x] / dx [dy/dx was done as shown above]
[Next, apply the Chain Rule to both derivatives]
= (d [cos^2 x] / d cos x) (d[cos x] / dx) - (d[sin^2 x] / d sin x) (d[sin x] / dx)
= (2cos x) (-sin x) - (2sin x) (cos x)
= -2sin x cos x - 2sin x cos x
= -4 sin x cos x

Answer 3

Use the multiplication formula for derivatives:
AB'+BA' (A = x) (B = Cos X); The derivative of cos x = -sin x
You would have: (x)(-sin x) + (cos x)(1)
Multiply and you will get: -x sin x + cos x (which can also be written as: cos x - x sin x)

Answer 4

dy/dx = 1 (cos x) + (- sin x) (x)
dy/dx = cos x - x (sin x)
d ² y / d x ² = - sin x - [ (1) sin x + cos x (x) ]
d ² y / d x ² = - sin x - sin x - (x) cos x
d ² y / d x ² = - 2 sin x - x (cos x)

Answer 5

dy/dx = sin(x)+x*cos(x)
d^2y/dx^2 = 2*cos(x)-x*sin(x)

Answer 6

Use the Product Rule.

If f(x) = x*cos(x), then...
f'(x) = cos(x) - x*sin(x).
f''(x) = df'/dx = -sin(x) - sin(x) - x*cos(x) = -2*sin(x) - x*cos(x).

:)