Question

Find the derivative of the following functions from first principle:
(i) -x
(ii) $(-x{)}^{-1}$
(iii) sin (x+1)
(iv) $cos(x-\frac{\pi }{8})$

Answer 1

(i) Here f(x)= -x

Then f(x+h)= -(x+h)

We know that:

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{-(x+h)-(-x)}{h}$

$={lim}_{h\to 0}\frac{-x-h+x}{h}$

$={lim}_{h\to 0}\frac{-h}{h}=-1$

(ii) Here f(x)= $(-x{)}^{-1}=-\frac{1}{x}$

Then $f(x+h)=-\frac{1}{x+h}$

We know that:

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{-\frac{1}{x+h}-f\left(x\right)}{h}$

$={lim}_{h\to 0}\frac{-x+x+h}{hx(x+h)}$

$={lim}_{h\to 0}\frac{h}{hx(x+h)}=\frac{1}{x^2}$

(iii) Here f(x)= sin (x+1)

Then f(x+h)= sin (x+h+1)

We know that

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{sin(x+h+1)-sin(x+1)}{h}$

$={lim}_{h\to 0}\frac{2cos\left(\frac{2x+h+2}{2}\right)sin\left(\frac{h}{2}\right)}{h}$

$={lim}_{h\to 0}\frac{cos(x+1+\frac{h}{2})sin\left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}$

$=cos(x+1)$

(iv) Here $f\left(x\right)=cos(x-\frac{\pi }{8})$

Then $f(x+h)=cos(x+h-\frac{\pi }{8})$

We know that

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{cos(x+h-\frac{\pi }{8})-cos(x-\frac{\pi }{8})}{h}$

$={lim}_{h\to 0}\frac{-2sin(x-\frac{\pi }{8}+\frac{h}{2})sin\left(\frac{h}{2}\right)}{h}$

$={lim}_{h\to 0}\frac{-sin(x-\frac{\pi }{8}+\frac{h}{2}).sin\left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}$

$=-sin(x-\frac{\pi }{8})$.