Question Number 24733 by chernoaguero@gmail.com last updated on 25/Nov/17
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{second}\:\mathrm{derivative}\:\mathrm{of} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\sqrt{\mathrm{5x}+\mathrm{9}} \\ $$$$\mathrm{find}\:\mathrm{f}^{''} \\ $$
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
![((Δy)/(Δx)) =(((√(5(x+△x)+9))−(√(5x+9)))/(△x)) =((5△x)/(△x[(√(5(x+△x)+9))+(√(5x+9))])) (dy/dx)=(5/(2(√(5x+9)))).](Q24774.png)
$$\frac{\Delta{y}}{\Delta{x}}\:\:=\frac{\sqrt{\mathrm{5}\left({x}+\bigtriangleup{x}\right)+\mathrm{9}}−\sqrt{\mathrm{5}{x}+\mathrm{9}}}{\bigtriangleup{x}} \\ $$$$\: \\ $$$$=\frac{\mathrm{5}\bigtriangleup{x}}{\bigtriangleup{x}\left[\sqrt{\mathrm{5}\left({x}+\bigtriangleup{x}\right)+\mathrm{9}}+\sqrt{\mathrm{5}{x}+\mathrm{9}}\right]} \\ $$$$\:\frac{{dy}}{{dx}}=\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{5}{x}+\mathrm{9}}}. \\ $$$$ \\ $$
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
![(d^2 y/dx^2 )=(5/2)lim_(△x→0) [(((1/(√(5(x+△x))))−(1/(√(5x+9))))/(△x))] =(5/2)lim_(△x→0) [(((√(5x+9))−(√(5(x+△x)))/((√(5x+9)) (√(5(x+△x)+9)) (△x)))] =(5/2)lim_(△x→0) [((−5△x)/(△x))]× lim_(△x→0) [(1/((√(5x+9)) (√(5(x+△x)+9))))]× lim_(△x→0) [(1/((√(5x+9))+(√(5(x+△x)+9))))] =(5/2)×(−5)×((1/(5x+9)))×((1/(2(√(5x+9))))) = −((25)/4)(5x+9)^(−3/2) .](Q24781.png)
$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{\mathrm{5}}{\mathrm{2}}\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\frac{\mathrm{1}}{\sqrt{\mathrm{5}\left({x}+\bigtriangleup{x}\right)}}−\frac{\mathrm{1}}{\sqrt{\mathrm{5}{x}+\mathrm{9}}}}{\bigtriangleup{x}}\right] \\ $$$$\:\:\:=\frac{\mathrm{5}}{\mathrm{2}}\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\sqrt{\mathrm{5}{x}+\mathrm{9}}−\sqrt{\mathrm{5}\left({x}+\bigtriangleup{x}\right.}}{\sqrt{\mathrm{5}{x}+\mathrm{9}}\:\sqrt{\mathrm{5}\left({x}+\bigtriangleup{x}\right)+\mathrm{9}}\:\left(\bigtriangleup{x}\right)}\right] \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{−\mathrm{5}\bigtriangleup{x}}{\bigtriangleup{x}}\right]× \\ $$$$\:\:\:\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\sqrt{\mathrm{5}{x}+\mathrm{9}}\:\sqrt{\mathrm{5}\left({x}+\bigtriangleup{x}\right)+\mathrm{9}}}\right]× \\ $$$$\:\:\:\:\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\sqrt{\mathrm{5}{x}+\mathrm{9}}+\sqrt{\mathrm{5}\left({x}+\bigtriangleup{x}\right)+\mathrm{9}}}\right] \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}×\left(−\mathrm{5}\right)×\left(\frac{\mathrm{1}}{\mathrm{5}{x}+\mathrm{9}}\right)×\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}{x}+\mathrm{9}}}\right) \\ $$$$=\:−\frac{\mathrm{25}}{\mathrm{4}}\left(\mathrm{5}{x}+\mathrm{9}\right)^{−\mathrm{3}/\mathrm{2}} \:. \\ $$
Commented by chernoaguero@gmail.com last updated on 26/Nov/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{thatz}\:\mathrm{correct} \\ $$