Find-the-second-derivative-of-f-x-5x-9-find-f-

Question Number 24733 by chernoaguero@gmail.com last updated on 25/Nov/17

Find the second derivative of

f (x) = \sqrt{5 x + 9}

find f^{^{″}}

Commented by chernoaguero@gmail.com last updated on 25/Nov/17

Using the first principle method

Answered by jota+ last updated on 25/Nov/17

\frac{Δ y}{Δ x} = \frac{\sqrt{5 (x + △ x) + 9} - \sqrt{5 x + 9}}{△ x}

= \frac{5 △ x}{△ x [\sqrt{5 (x + △ x) + 9} + \sqrt{5 x + 9}]}

\frac{d y}{d x} = \frac{5}{2 \sqrt{5 x + 9}} .

Commented by chernoaguero@gmail.com last updated on 25/Nov/17

l mean the second derivative

now continue from ur ans using

the first principal too

Commented by ajfour last updated on 25/Nov/17

\frac{d^{2} y}{{d x}^{2}} = \frac{5}{2} \lim_{△ x \to 0} [\frac{\frac{1}{\sqrt{5 (x + △ x)}} - \frac{1}{\sqrt{5 x + 9}}}{△ x}]

= \frac{5}{2} \lim_{△ x \to 0} [\frac{\sqrt{5 x + 9} - \sqrt{5 (x + △ x}}{\sqrt{5 x + 9} \sqrt{5 (x + △ x) + 9} (△ x)}]

= \frac{5}{2} \lim_{△ x \to 0} [\frac{- 5 △ x}{△ x}] \times

\lim_{△ x \to 0} [\frac{1}{\sqrt{5 x + 9} \sqrt{5 (x + △ x) + 9}}] \times

\lim_{△ x \to 0} [\frac{1}{\sqrt{5 x + 9} + \sqrt{5 (x + △ x) + 9}}]

= \frac{5}{2} \times (- 5) \times (\frac{1}{5 x + 9}) \times (\frac{1}{2 \sqrt{5 x + 9}})

= - \frac{25}{4} {(5 x + 9)}^{- 3 / 2} .

Commented by chernoaguero@gmail.com last updated on 26/Nov/17

Thank you sir thatz correct