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): $\frac{x}{1+tan\,x}$

Answer 1

Let f(x)=$\frac{x}{1+tan\,x}$
f'(x)=(1+tanx)$\frac{d}{dx}$(x)-x$\frac{d}{dx}$$\frac{(1+tan\,x)}{(1+tan\,x)^2}$
f'(x)=(1+tanx) -x.$\frac{d}{dx}$$\frac{(1+tan\,x)}{(1+tan\,x)^2}$ ..(i)
Let g(x)=1+tan x. Accordingly, g(x+h)=1+tan (x + h).
By first principle
$g'$(x)=$\lim_{h\rightarrow 0} \frac{g(x+h)-g(x)}{h}$
=$\lim_{h\rightarrow 0}$ [$\frac{1+tan(x+h)-1-tan\,x}{h}$]
=$\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{sin(x+h)}{cos(x+h)}-\frac{sin\,x}{cos\,x}$]
=$\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{sin(x+h)cos\,x-sin\,xcos(x+h)}{cos(x+h)cos\,x}$]
=$\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{sin(x+h-x)}{cos(x+h)cos\,x}$]
=$\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{sin\,h}{cos(x+h)cos\,x}$]
=($\lim_{h\rightarrow 0}$ $\frac{sin\,h}{h}$).($\lim_{h\rightarrow 0}$$\frac{1}{cos(x+h)cos\,x}$
=1x$\frac{1}{cos^2x}$ = sec²x
$\Rightarrow$(1+tanx) = sec²x ...(ii)
From (i) and (ii), we obtain
$f'$(x)=$\frac{1+tan\,x-xsec^2x}{(1+tan\,x)^2}$