Question

compute the first three derivatives of (f(x)=-\frac{1}{x}), and then state a formula for (f^{(n)}(x)). a. (f(x)=) b. (f(x)=) c. (f(x)=) d. (f^{(n)}(x)=)

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=-\frac{1}{x}$ as $f(x)=-x^{-1}$.

Step2: Find the first - derivative

Use the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$. So, $f'(x)=(-1)\times(-1)x^{-1 - 1}=x^{-2}=\frac{1}{x^{2}}$.

Step3: Find the second - derivative

Differentiate $f'(x)=x^{-2}$ again using the power rule. $f''(x)=(-2)x^{-2 - 1}=-2x^{-3}=-\frac{2}{x^{3}}$.

Step4: Find the third - derivative

Differentiate $f''(x)=-2x^{-3}$ using the power rule. $f'''(x)=(-2)\times(-3)x^{-3 - 1}=6x^{-4}=\frac{6}{x^{4}}$.

Step5: Find the n - th derivative formula

By observing the pattern:
$f'(x)=1!x^{-2}$, $f''(x)=-2!x^{-3}$, $f'''(x)=3!x^{-4}$.
So, $f^{(n)}(x)=(-1)^{n + 1}n!x^{-(n + 1)}=\frac{(-1)^{n+1}n!}{x^{n + 1}}$.

Answer:

a. $f'(x)=\frac{1}{x^{2}}$
b. $f''(x)=-\frac{2}{x^{3}}$
c. $f'''(x)=\frac{6}{x^{4}}$
d. $f^{(n)}(x)=\frac{(-1)^{n + 1}n!}{x^{n+1}}$