Nth Derivative of x³ cosx | Leibnitz Theorem Solved Example
Step-by-Step Solution
Step 1: Recall Leibnitz Theorem
For two functions u(x) and v(x), the nth derivative of their product is given by:
(u v)⁽ⁿ⁾ = Σ (n choose k) * u⁽ⁿ⁻ᵏ⁾ * v⁽ᵏ⁾, k = 0 to n
Step 2: Identify u(x) and v(x)
Here, we have u(x) = x³ and v(x) = cos(x).
Step 3: Compute derivatives of u(x)
x³ derivatives:
- u' = 3x²
- u'' = 6x
- u''' = 6
- u⁽⁴⁾ = 0 (and all higher derivatives are 0)
Step 4: Compute derivatives of v(x)
cos(x) derivatives follow a cycle of 4:
- v = cos(x)
- v' = -sin(x)
- v'' = -cos(x)
- v''' = sin(x)
- v⁽⁴⁾ = cos(x), and so on.
Step 5: Apply Leibnitz Theorem
The nth derivative of x³ cos(x) is:
(x³ cosx)⁽ⁿ⁾ = Σ (n choose k) * u⁽ⁿ⁻ᵏ⁾ * v⁽ᵏ⁾
Since derivatives of x³ vanish after 3, the sum only includes k = n, n-1, n-2, n-3.
Step 6: Write explicit formula
(x³ cosx)⁽ⁿ⁾ = (n choose 0) * x³ * v⁽ⁿ⁾ + (n choose 1) * 3x² * v⁽ⁿ⁻¹⁾ + (n choose 2) * 6x * v⁽ⁿ⁻²⁾ + (n choose 3) * 6 * v⁽ⁿ⁻³⁾
All terms for u⁽⁴⁾ and higher vanish because u⁽⁴⁾ = 0.
Step 7: Substitute derivatives of cosx for required n
Use the cycle: cos(x), -sin(x), -cos(x), sin(x), … to evaluate v⁽ⁿ⁾, v⁽ⁿ⁻¹⁾, etc.
Step 8: Final Expression
Thus, the nth derivative of x³ cosx can be written explicitly using the formula in Step 6 and substituting the derivatives of cosx based on the cycle.