Question

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed nonzero constants and m and n are integers): $(x+sec\,x)(x-tan\,x)$

Answer 1

Let f(x) = (x+secx) (x−tan x)
By product rule,
f'(x) = (x + secx)$\frac{d}{dx}$(x−tanx)+(x−tanx)$\frac{d}{dx}$(x + sec x)
=(x+ secx) [$\frac{d}{dx}$(x)-$\frac{d}{dx}$tan x]+(x-tan.x) [$\frac{d}{dx}$(x)+$\frac{d}{dx}$sec x]
=(x + secx)[1- $\frac{d}{dx}$ tan x]+(x- tan x)[1 + $\frac{d}{dx}$ sec x] ...(i)
Let $f_1$ (x)= tan x,$f_2$(x) = secx
Accordingly, f1(x+h) = tan(x+h) and f2 (x+h) = sec(x+h)
$f'$1(x) = $\lim_{h\rightarrow 0}$($\frac{f_1(x+h)-f_1(x)}{h}$)
= $\lim_{h\rightarrow 0}$ ($\frac{tan(x+h)-tan\,x}{h}$)
= $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{sin(x+h)}{cos\,x}-\frac{sin\,x}{cos\,x}$]
= $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[ $\frac{sin\,h}{cos(x+h)cos\,x}$]
=($\lim_{h\rightarrow 0}$ $\frac{sin\,h}{h}$). ($\lim_{h\rightarrow 0}$ $\frac{1}{cos(x+h)}$ cos x)
= 1\times$$\frac{1}{cos^2x} = sec2x
⇒$\frac{d}{dx}$ tan x = sec2x ...(ii)
$f'$2(x) = $\lim_{h\rightarrow 0}$($\frac{f_2(x+h)-f_2(x)}{h}$)
= $\lim_{h\rightarrow 0}$($\frac{sec(x+h)-sec\,x}{h}$)
= $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[ $\frac{1}{cos(x+h)}-\frac{1}{cos\,x}$]
= $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{cos\,x-cos(x+4)}{cos(x+4)cos\,x}$]
=$\frac{1}{cos\,x}$ $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[[sin($\frac{2x+h}{2}$) {$\frac{sin\frac{h}{2}}{\frac{h}{2}}$}]
secx. tanx ...(iii)
From (i), (ii), and (iii), we obtain
$f'$(x) = (x+sec x)(1-sec2x) + (x-tan x)(1+sec x tan x)