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

$$\mathrm{Find}\mathrm{the}\mathrm{second}\mathrm{derivative}\mathrm{of}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{5\mathrm{x}+9}$$$$\mathrm{find}{\mathrm{f}}^{{}^{″}}$$

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

$$\mathrm{Using}\mathrm{the}\mathrm{first}\mathrm{principle}\mathrm{method}$$

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

$$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{\sqrt{5(x+△x)+9}-\sqrt{5x+9}}{△x}$$$$$$$$=\frac{5△x}{△x[\sqrt{5(x+△x)+9}+\sqrt{5x+9}]}$$$$\frac{dy}{dx}=\frac{5}{2\sqrt{5x+9}}.$$$$$$

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

$$\mathrm{l}\mathrm{mean}\mathrm{the}\mathrm{second}\mathrm{derivative}$$$$\mathrm{now}\mathrm{continue}\mathrm{from}\mathrm{ur}\mathrm{ans}\mathrm{using}$$$$\mathrm{the}\mathrm{first}\mathrm{principal}\mathrm{too}$$

Commented by ajfour last updated on 25/Nov/17

$$\frac{d^{2}y}{{dx}^2}=\frac{5}{2}\underset{△x\to 0}{\lim }\left[\frac{\frac{1}{\sqrt{5(x+△x)}}-\frac{1}{\sqrt{5x+9}}}{△x}\right]$$$$=\frac{5}{2}\underset{△x\to 0}{\lim }\left[\frac{\sqrt{5x+9}-\sqrt{5(x+△x}}{\sqrt{5x+9}\sqrt{5(x+△x)+9}(△x)}\right]$$$$=\frac{5}{2}\underset{△x\to 0}{\lim }\left[\frac{-5△x}{△x}\right]\times$$$$\underset{△x\to 0}{\lim }\left[\frac{1}{\sqrt{5x+9}\sqrt{5(x+△x)+9}}\right]\times$$$$\underset{△x\to 0}{\lim }\left[\frac{1}{\sqrt{5x+9}+\sqrt{5(x+△x)+9}}\right]$$$$=\frac{5}{2}\times (-5)\times \left(\frac{1}{5x+9}\right)\times \left(\frac{1}{2\sqrt{5x+9}}\right)$$$$=-\frac{25}{4}{(5x+9)}^{-3/2}.$$

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

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}\mathrm{thatz}\mathrm{correct}$$

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