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Mathematics Question on Limits and derivations

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): x1+tan x\frac{x}{1+tan\,x}

Answer

Let f(x)=x1+tan x\frac{x}{1+tan\,x}
f'(x)=(1+tanx)ddx\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′g'(x)=lim⁡h→0g(x+h)−g(x)h\lim_{h\rightarrow 0} \frac{g(x+h)-g(x)}{h}
=lim⁡h→0\lim_{h\rightarrow 0} [1+tan(x+h)−1−tan xh\frac{1+tan(x+h)-1-tan\,x}{h}]
=lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[sin(x+h)cos(x+h)−sin xcos x\frac{sin(x+h)}{cos(x+h)}-\frac{sin\,x}{cos\,x}]
=lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[sin(x+h)cos x−sin xcos(x+h)cos(x+h)cos x\frac{sin(x+h)cos\,x-sin\,xcos(x+h)}{cos(x+h)cos\,x}]
=lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[sin(x+h−x)cos(x+h)cos x\frac{sin(x+h-x)}{cos(x+h)cos\,x}]
=lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[sin hcos(x+h)cos x\frac{sin\,h}{cos(x+h)cos\,x}]
=(lim⁡h→0\lim_{h\rightarrow 0} sin hh\frac{sin\,h}{h}).(\lim_{h\rightarrow 0}$$\frac{1}{cos(x+h)cos\,x}
=1x1cos2x\frac{1}{cos^2x} = sec2x
⇒\Rightarrow(1+tanx) = sec2x ...(ii)
From (i) and (ii), we obtain
f′f'(x)=1+tan x−xsec2x(1+tan x)2\frac{1+tan\,x-xsec^2x}{(1+tan\,x)^2}