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)