## Calculus (3rd Edition)

$f^{(k)}(x)=0$ implies that the derivative eventually becomes zero for all $k$ values. a) As there is a $x^{-1}$ term, by the power rule, the power of $x$ becomes more and more negative when we differentiate again and again. Therefore, the power can never be 0. b) $f'(x)=3x^{2}$ $f''(x)=6x$ $f^{(3)}(x)=6x^{0}=6$ $f^{(4)}(x)=0$ Differentiating $0$ again and again, we get 0. Therefore, the condition is satisfied. c) While differentiating $\sqrt x=x^{1/2}$ again and again, as the power of $x$ never becomes 0, $f^{(k})(x)$ will never be equal to 0. d) The power of $x$ will become $6-6=0$ after differentiating 6 times. If we differentiate after that, we get $f^{(7)}(x)=0$, etc. Therefore, the condition is satisfied. e) While differentiating $x^{9/5}$ again and again, as the power of $x$ never becomes 0, $f^{(k})(x)$ will never be equal to 0. f) $f'(x)= 4x+15x^{4}$ $f''(x)=4+60x^{3}$ $f'''(x)=0+180x^{2}$ $f^{(4)}(x)=360x$ $f^{(5)}(x)=360$ $f^{(6)}(x)=0$ The condition is satisfied.