f(x)=√(x+9)
f(x+h) =√(x+h+9)
[tex] f'(x) = \lim_{h \to \\0} \frac{\sqrt{x+9+h}-\sqrt{x+9}}{h} =
\\ \\ = \lim_{h \to \\0} \frac{(\sqrt{x+9+h}-\sqrt{x+9})*(\sqrt{x+9+h}+\sqrt{x+9})}{h(\sqrt{x+9+h}+\sqrt{x+9})} =
\\ \\= \lim_{h \to \\0} \frac{(\sqrt{x+9+h})^{2}-(\sqrt{x+9})^{2}}{h(\sqrt{x+9+h}+\sqrt{x+9})}=
\\ \\= \lim_{n \to \\0} \frac{x+9+h-x-9}{h(\sqrt{x+9+h}+\sqrt{x+9})}=\lim_{n \to \\0} \frac{h}{h(\sqrt{x+9+h}+\sqrt{x+9})}=\lim_{n \to \\0} \frac{1}{(\sqrt{x+9+h}+\sqrt{x+9})} =
\\ \\= \frac{1}{(\sqrt{x+9+0}+\sqrt{x+9})} = [/tex]
[tex] =\frac{1}{(\sqrt{x+9}+\sqrt{x+9})} = \frac{1}{2\sqrt{x+9}} [/tex]
[tex] f'(x)= \frac{1}{2\sqrt{x+9}}
\\ \\ f'(a)=\frac{1}{2\sqrt{a+9}}
\\ \\a=7
\\ \\f'(7)= \frac{1}{2\sqrt{7+9}} = \frac{1}{2*4} =\frac{1}{8}
\\ \\f'(7)= \frac{1}{8} [/tex]
