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):

$$\dfrac{x}{1+tanx}$$

Asked by Abhisek | 1 year ago |  118

##### Solution :-

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

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