Question

Find f' f(x)=sqr root(x+9), a=7use the definition lim f(x+h)-f(x)/h

Answer 1

f(x)=√(x+9)

f(x+h) =√(x+h+9)

$f'(x) = \lim_{h \to \\0} \frac{\sqrt{x+9+h}-\sqrt{x+9}}{h} = \\ \\ = \lim_{h \to \\0} \frac{(\sqrt{x+9+h}-\sqrt{x+9})*(\sqrt{x+9+h}+\sqrt{x+9})}{h(\sqrt{x+9+h}+\sqrt{x+9})} = \\ \\= \lim_{h \to \\0} \frac{(\sqrt{x+9+h})^{2}-(\sqrt{x+9})^{2}}{h(\sqrt{x+9+h}+\sqrt{x+9})}= \\ \\= \lim_{n \to \\0} \frac{x+9+h-x-9}{h(\sqrt{x+9+h}+\sqrt{x+9})}=\lim_{n \to \\0} \frac{h}{h(\sqrt{x+9+h}+\sqrt{x+9})}=\lim_{n \to \\0} \frac{1}{(\sqrt{x+9+h}+\sqrt{x+9})} = \\ \\= \frac{1}{(\sqrt{x+9+0}+\sqrt{x+9})} =$

$=\frac{1}{(\sqrt{x+9}+\sqrt{x+9})} = \frac{1}{2\sqrt{x+9}}$

$f'(x)= \frac{1}{2\sqrt{x+9}} \\ \\ f'(a)=\frac{1}{2\sqrt{a+9}} \\ \\a=7 \\ \\f'(7)= \frac{1}{2\sqrt{7+9}} = \frac{1}{2*4} =\frac{1}{8} \\ \\f'(7)= \frac{1}{8}$