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): cosec x cot x

Asked by Abhisek | 1 year ago |  68

##### Solution :-

Let f(x) = cosec x cot x

By Leibnitz product rule,

f’(x) = cosec x(cot x)’+cot x(cosec x)’ ….(1)

Let f1(x) = cot x.

Accordingly, f1(x + h) = cot (x + h)

By first principle,

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

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

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

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

$$\dfrac{-1}{sinx}.1.( \lim\limits_{h \to 0}\dfrac{1}{sin(x+0)})$$

$$\dfrac{-1}{sin^2x}$$

cosec2x=− (cotx)'= - cosxec2x……(2)

Now, let f2(x) = cosec x.

Accordingly, f2(x + h)  = cosec(x + h)

By first principle,

$$f_2'(x)=\lim\limits_{h \to 0}\dfrac{f_2(x+h)-f_2(x)}{h}$$

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

$$\lim\limits_{h \to 0}\dfrac{1}{h}(\dfrac{sinx-sin(x+h)}{sinxsin(x+h)})$$

$$\dfrac{-1}{sinx}.1.\dfrac{(\dfrac{2x+h}{2})}{sin(x+0)}$$

$$\dfrac{-1}{sinx}.\dfrac{cosx}{sinx}$$

(cosec x)’=-cosec x. cot x

From (1), (2), and (3), we obtain

f'(x) = cosec x(-cosec2x) +cot x(-cosec x cot x)

= -cosec3 x-cot2 x cosec x

Answered by Pragya Singh | 1 year ago

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