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