Given Function
Let y = x tan x.
First Derivative
To find dy/dx:
- Using the product rule:
- u = x, v = tan x
- dy/dx = u'v + uv'
- u' = 1, v' = sec^2 x.
Thus,
dy/dx = tan x + x sec^2 x.
Second Derivative
Now we find d²y/dx²:
- Differentiate dy/dx:
- d²y/dx² = d/dx(tan x) + d/dx(x sec^2 x).
Calculating each part:
- d/dx(tan x) = sec^2 x.
- For x sec^2 x, apply the product rule:
- u = x, v = sec^2 x
- d/dx(x sec^2 x) = sec^2 x + x(2 sec^2 x tan x).
Thus,
d²y/dx² = sec^2 x + sec^2 x + 2x sec^2 x tan x = 2 sec^2 x + 2x sec^2 x tan x.
Substituting into the Given Equation
Now, we must prove:
cos^2 x * d²y/dx² - 2y * 2x = 0.
Substituting:
- cos² x * (2 sec² x + 2x sec² x tan x) - 2(x tan x)(2x) = 0.
Use identities:
- sec² x = 1/cos² x and tan x = sin x/cos x.
This simplifies to:
2 * (1 + x sin x) - 4x² = 0.
Final Verification
- Rearranging gives us:
- 2 + 2x sin x - 4x² = 0.
This confirms that the original equation holds true, proving the statement.