Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (x + cos x) (x – tan x)

Asked by Abhisek | 1 year ago |  117

##### Solution :-

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,

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

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

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

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

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

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

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

$$\dfrac{1}{cos^2x}$$

= sec2x ..................…(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

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