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): (4x+5sinx)(3x+7cosx)

Answer 1

Let $f'$(x)=(x+cos x) (x-tan x)
By product rule,
f'(x) = (x + cos x)$\frac{d}{dx}$(x - tan x)+(x - tan x) $\frac{d}{dx}$(x + cos x)
= (x + cos x) [ $\frac{d}{dx}$(x) - $\frac{d}{dx}$(tan x)] + (x-tan x)(1— sin x)
= (x + cos x)[1-$\frac{d}{dx}$tan x]+(x-tan x) (1-sin x) ...(i)
Let g(x) = tan x. Accordingly, g(x+h) = tan(x+h)
By first principle.
g'(x)= $\lim_{h\rightarrow 0}$ $\frac{g(x+h)-g(x)}{h}$
= \lim_{h\rightarrow 0}$$\frac{tan(x+h)-tan\,x}{h}
= $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{sin(x+h)}{cos(x+h)}-sin\,x\frac{h}{cos\,x}$]
= $\lim_{h\rightarrow 0}$ $\frac{1}{h}$ [$\frac{sin(x+h)cos\,x-sin\,xcos(x+h)}{cos(x+h)cos\,x}$]
=\frac{1}{cos\,x}$$\lim_{h\rightarrow 0} $\frac{1}{h}$[$\frac{sin(x+h-x)}{cos(x+h)}$]
=\frac{1}{cos\,x}$$\lim_{h\rightarrow 0} $\frac{1}{h}$[$\frac{sin\,h}{cos(x+h)}$]
=$\frac{1}{cos\,x}$ ($\lim_{h\rightarrow 0}$ $sin\frac{h}{h}$).($\lim_{h\rightarrow 0}$ $\frac{1}{cos(x+h)}$)
=$\frac{1}{cos\,x}$.1.$\frac{1}{cos\,x}$(x+0)
=$\frac{1}{cos^2x}$ = sec2x ...(ii)
Therefore, from (i) and (ii), we obtain
$f'$(x)=(x+cosx)(1-sec2x)+(x-tan x) (1-sin x)
=(x+cos.x) (-tan2x)+(x−tan x)(1−sin x)
=−tan2x(x+cos x)+(x−tan x)(1−sin x)