Question

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed nonzero constants and m and n are integers): cosec x cot x

Answer 1

Let f(x) =cosecx cotx
By Leibnitz product rule,
f'(x)=cosec x(cotx)'+cot x(cosec x)' ...(1)
Let f1(x) = cotx. Accordingly, f1(x+h) = cot(x+h)
By first principle,
f'1(x) = $\lim_{h\rightarrow 0}$ f2(x+h) − f1(x)/h
= $\lim_{h\rightarrow 0}$ cot(x+h) - $\frac{cot\,x}{h}$
= $\lim_{h\rightarrow 0}$ $\frac{1}{h}$( $\frac{cos(x+h)}{sin(x+h)}-\frac{cos\,x}{sin\,x}$)
= $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{sin\,xcos(x+h)-cos\,xsin(x+h)}{sin\,xsin(x+h)}$]
= $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{sin(x-x-h)}{sin\,xsin(x+h)}$]
= $\frac{1}{sin\,x}$. \lim_{h\rightarrow 0}$$\frac{1}{h}[ $\frac{sin(-h)}{sin(x+h}$]
=-$\frac{1}{sin\,x}$.1.($\frac{1}{sin}$(x+0))
=$-\frac{1}{sin^2x}$
=-cosec2x
∴(cotx)' = -cosec2x ...(2)
Now, let f(x) = cosec x. Accordingly, f2(x+h) = cosec(x+h)
By first principle,
f'2(x) = $\lim_{h\rightarrow 0}$ $\frac{f_2(x+h)-f_2(x)}{h}$
= $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[cosec(x+h) - cosecx]
=$\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{2}{sin(x+h)}-\frac{1}{sin\,x}$]
= $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{1}{sin\,x}$ - $\frac{sin(x+h)}{sin\,xsin(x+h)}$]
=$\frac{1}{sin\,x}$ $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[2cos ($\frac{x+x+h}{2}$) sin($\frac{x-x-h}{2}$)]
=$\frac{1}{sin\,x}$ $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[2cos ($\frac{2x+h}{2}$) sin($\frac{-h}{2}$)]
=$\frac{1}{sin\,x}$ $\lim_{h\rightarrow 0}$ $\frac{1}{h}$[$\frac{-sin\frac{h}{2}}{\frac{h}{2}}$.cos($\frac{2x+h}{2}$)]
=-$\frac{1}{sin\,x}$ $\lim_{h\rightarrow 0}$ $\frac{sin\frac{h}{2}}{\frac{h}{2}}$ . $\lim_{h\rightarrow 0}$ $\frac{cos\frac{2x+h}{2}}{sin(x+h)}$
=-$\frac{1}{sin\,x}$.1.1.$\frac{cos(\frac{2x+0}{2})}{sin(x+0)}$=$-\frac{1}{sin\,x}.\frac{cos\,x}{sin\,x}$
=-cosecx.cotx
∴(cosecx)' = -cosecx cotx ...(3)
From (1), (2), and (3), we obtain
f'(x) = cosecx (-cosec2x) + cotx(-cosecx cotx)
=-cosec3x-cot2xcosecx