Let f(x) = \( \dfrac{x}{1+tanx}\)

\( f'(x)=\dfrac{(1+tanx)-x\dfrac{d}{dx}(1+tanx)}{(1+tanx)^2}\)

Let g(x) = 1 + tan x. .Accordingly, g(x + h) = 1 + tan(x+h). By first principle,

\( g'(x)=\lim\limits_{h \to 0}\dfrac{g(x+h)-g(x)}{h}\)

\(\lim\limits_{h \to 0}[\dfrac{1+tan(x+h)-1-tan(x)}{h}]\)

\( \lim\limits_{h \to 0}[\dfrac{1}{h}[\dfrac{sin(x+h)}{cos(x+h)}-\dfrac{sinx}{cosx}]\)

\( \lim\limits_{h \to 0}[\dfrac{1}{h}[\dfrac{sin(x+h-x)}{cos(x+h)cosx}]\)

\( \lim\limits_{h \to 0}[\dfrac{1}{h}[\dfrac{sin h}{cos(x+h)cosx}]\)

\(( \lim\limits_{h \to 0}\dfrac{sinh}{h}). ( \lim\limits_{h \to 0}\dfrac{1}{cos(x+h)cosx})\)

\( 1\times \dfrac{1}{cos^2}=sec^2x\)

\( \dfrac{d}{dx}(1+tan^2x)=sec^2x\) ..........(ii)

From (i) and (ii),we obtain

\( f'(x)=\dfrac{1+tanx-xsec^2x}{(1+tan)^2}\)

Answered by Pragya Singh | 1 year ago