Answer to Question #128705 in Calculus for Onyinye

Question

Using the first principle, find the derivative of the function f (x) = 1 . (Show all workings and √x+1
state clearly, any theorem used)

Answer 1

It follows from first principles that the derivative of a function $f(x)$ is

$$f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$$

In our case, for $f(x)=1$

$$f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim\limits_{h\to0}\frac{(1)-1}{h}=\\[0.3cm] =\lim\limits_{h\to0}\left(\frac{0}{h}\right)=0\to\boxed{f(x)=1\to f'(x)=0}$$

For $f(x)=\sqrt{x+1}$

$$f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim\limits_{h\to0}\frac{\sqrt{x+1+h}-\sqrt{x+1}}{h}=\\[0.3cm] =\lim\limits_{h\to0}\frac{\left(\sqrt{x+1+h}-\sqrt{x+1}\right)\left(\sqrt{x+1+h}+\sqrt{x+1}\right)}{h\left(\sqrt{x+1+h}+\sqrt{x+1}\right)}=\\[0.3cm] =\lim\limits_{h\to0}\frac{\left(x+1+h\right)-\left(x+1\right)}{h\left(\sqrt{x+1+h}+\sqrt{x+1}\right)}=\\[0.3cm] =\lim\limits_{h\to0}\frac{h}{h\left(\sqrt{x+1+h}+\sqrt{x+1}\right)}=\\[0.3cm] =\lim\limits_{h\to0}\frac{1}{\left(\sqrt{x+1+h}+\sqrt{x+1}\right)}=\frac{1}{2\sqrt{x+1}}\\[0.3cm] \boxed{f(x)=\sqrt{x+1}\to f'(x)=\frac{1}{2\sqrt{x+1}}}$$