Let f(x) = (x + cos x) (x – tan x) By product rule,

\( f'(x)=(x+cosx)\dfrac{d}{dx}(x-tanx)\)

\( +(x-tanx)\dfrac{d}{dx}(x+cosx)\)

\( =(x+cosx)[\dfrac{d}{dx}(x)-\dfrac{d}{dx}(tanx)]\)

\( +(x-tanx)(1-sinx)\)

\( (x+cosx)[1-\dfrac{d}{dx}(tanx)]\)

\( +(x-tanx)(1-sinx)\) ............(i)

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

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

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

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

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

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

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

\( \dfrac{1}{cosx}.1.(\dfrac{1}{cos(x+0)})\)

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

= sec^{2}x ..................…(ii)

Therefore, from (i) and (ii) ,We obtain

\( f'(x)=(x+cosx)(1-sec^2x)+(x-tanx)(1-sinx)\)

\( (x+cosx)(-tan^2x)+(x-tanx)(1-sinx)\)

\( -tan^2x(x+cosx)+(x-tanx)(1-sinx)\)

Answered by Pragya Singh | 1 year ago