Higher Order Derivatives — Question 8

Question 8 - Solution

We are given: $$f(x) = \frac{1}{x} = x^{-1}$$

(a) Compute derivatives:

$$\begin{align*} f'(x) &= \frac{d}{dx}(x^{-1}) = -x^{-2} \\ f''(x) &= \frac{d}{dx}(-x^{-2}) = 2x^{-3} \\ f^{(3)}(x) &= \frac{d}{dx}(2x^{-3}) = -6x^{-4} \\ f^{(4)}(x) &= \frac{d}{dx}(-6x^{-4}) = 24x^{-5} \end{align*}$$

(b) General Pattern:

Observe the signs and coefficients:

$$f^{(n)}(x) = (-1)^n \cdot n! \cdot x^{-(n+1)}$$

Answer: $\boxed{f^{(n)}(x) = (-1)^n \cdot n! \cdot x^{-(n+1)}}$

(c) Evaluate $f^{(6)}(x)$:

Use the formula: $$f^{(6)}(x) = (-1)^6 \cdot 6! \cdot x^{-7} = 720x^{-7}$$

Answer: $\boxed{f^{(6)}(x) = \frac{720}{x^7}}$