Question

Find the derivative of the following functions:
(i) sin x cos x 
(ii) sec x 
(iii) 5sec x+4cos x
(iv) cosec x 
(v) 3cot x+5cosec x
(vi) 5sin x-6cos x+7 
(vii) 2tan x-7sec x

Answer 1

(i) Letf (x) = sin x cos x. Accordingly, from the first principle,
f'(x)= \(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{sin(x+h)cos(x+h)-sin\,xcos\,x}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{1}{2h}\)[sin 2(x+h)-sin 2x]
= \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[ cos\(\frac{4x+2h}{2}\) sin \(\frac{2h}{2}\)]
= \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)cos(2x+h) \(\lim_{h\rightarrow 0}\) \(\frac{sin\,h}h{}\)
= cos(2x+0).1
= cos 2.x

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(ii) Letf (x) = sec x. Accordingly, from the first principle,
f'(x)= \(\lim_{h\rightarrow 0}\)\(\frac{f(x+h)-f(x)}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{sec(x+h)-sec(x)}{h}\)
= \(\lim_{h\rightarrow 0}\) 1h[\(\frac{1}{cos(x+h)}-\frac{1}{cos\,x}\)]
= \(\lim_{h\rightarrow 0}\) \(\frac{1}{h_1}\)[ \(\frac{cos\,x-cos(x+h)}{cos\,xcos(x+h)}\)]
=\(\frac{1}{cos\,x}\) \(\lim_{h\rightarrow 0}\)[-2sin (x+x+\(\frac{h}{2}\)) sin(x-x-\(\frac{h}{2}\)) / cos(x+h)]
=\(\frac{1}{cos\,x}\) \(\lim_{h\rightarrow 0}\)[\(\frac{-2sin(\frac{2x+h}{2})sin(\frac{h}{2})}{cos(x+h)}\)]
=\(\frac{1}{cos\,x}\)\(\lim_{h\rightarrow 0}\)[-2sin (\(\frac{2x+h}{2}\)) sin(\(\frac{h}{2}\))cos(x+h)]
=\(\frac{1}{cos\,x}\) \(\lim_{\frac{h}{2}\rightarrow 0}\) \(\frac{sin\frac{h}{2}}{\frac{h}{2}}\).\(\lim_{h\rightarrow 0}\) \(\frac{sin(\frac{2x+h}{2})}{cos(x+h)}\)
=\(\frac{1}{cosx}\) .1.\(\frac{sin\,x}{cos\,x}\)
= sec x tan x

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(iii) Letf (x) = 5 sec x + 4 cos x. Accordingly, from the first principle,
f'(x)= \(\lim_{h\rightarrow 0}\)\(\frac{f(x+h)-f(x)}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{5\,sec(x+h)+4cos(x+h)-[5\,sec\,x+cos\,x]}{h}\)
= 5\(\lim_{h\rightarrow 0}\)\(\frac{sec(x+h)-sec(x)}{h}\) +4 \(\lim_{h\rightarrow 0}\)\(\frac{cos(x+h)-cos(x)}{h}\)
= 5\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) \([\frac{cos\,x-cos(x+h)}{cos\,xcos(x+h)}]\) +4\(\lim_{h\rightarrow 0}\) [cosxcosh–sinxsinh −cosx]
=\(\frac{5}{cosx}\)\(\lim_{h\rightarrow 0}\)\([\frac{sin\frac{2x+h}{2}.sin(\frac{sin(\frac{-h}{2})}{\frac{h}{2}})}{cos(x+h)}]\)
=\(\frac{5}{cosx}\) [\(\lim_{h\rightarrow 0}\) \(\frac{sin\frac{2x+h}{2}}{cos(x+h)}\) . \(\lim_{h\rightarrow 0}\)\(\frac{sin(-\frac{h}{2})}{\frac{h}{2}}\)] - 4sinx
=\(\frac{5}{cosx}\) .\(\frac{sin\,x}{cos\,x}\).1-4sinx
=5sec x tan x-4sin x

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(iv) Let f (x) = cosec x. Accordingly, from the first principle,
f'(x) = \(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
f'(x)=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[\(cosec(x+h)-cosec\,x\)]
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)\([\frac{1}{sin(x+h)}-\frac{1}{sin\,x}]\)
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[\(\frac{sin\,x-sin(x+h)}{sin(x+h)sin\,x}\)]
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) [2cos(\(\frac{x+x+h}{2}\)). \(\frac{sin(\frac{x-x-h}{2})}{sin(x+h)sin\,x}\)] 
=\(\lim_{h\rightarrow 0}\) [\(\frac{cos\frac{2x+h}{2}.sin(\frac{h}{2})}{sin(x+h)sin\,x}\)]
=\(\lim_{h\rightarrow 0}\) \([\frac{-cos(\frac{2x+h}{2}).(\frac{sin\frac{h}{2}}{\frac{h}{2}})}{sin(x+h)sin\,x}]\)
=(\(\frac{-cos\,x}{sinx\.sinx}\)).1
=-cosecx cotx

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(v) Let f (x) = 3cot x + 5cosec x. Accordingly, from the first principle,
f'(x) = \(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{3\,cot(x+h+5cosec(x+h))-3\,cot\,x-5\,cosec\,x}{h}\)
=3\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[cot (x+4)-cot.x]+5\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) [cosec (x+h) - cosec x] ...(1)
Now, \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) [cot (x+h)-cot x]
=\(\lim_{h\rightarrow 0}\)\(\frac{1}{h}\) \([\frac{cos(x+h)}{sin(x+h)}-\frac{cos\,x}{sin\,x}]\)
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) [\(\frac{sin(x-x-h)}{sin\,xsin(x+h)}\)]
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[\(\frac{sin(-h)}{sin\,x\,sin(x+4)}\)]
=-(\(\lim_{h\rightarrow 0}\) \(\frac{sin\,h}{h}\)).(\(\lim_{h\rightarrow 0}\) \(\frac{1}{sin\,x}\)-sin(x+h)
=-1. \(\frac{1}{sin\,x}\) sin(x+0) = \(-\frac{1}{sin^x}\) = -cosec²x ...(2)
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) [cosec(x+h)-cosecx]
=\(\lim_{h\rightarrow 0}\)\(\frac{1}{h}\)[\(\frac{1}{sin(x+h)}-\frac{1}{sin\,x}\)]
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[\(\frac{sin\,x-sin(x+h)}{sin(x+h)sin\,x}\)]
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) [\(\frac{2cos(x + x +\frac{1}{2}). sin (x-x-\frac{h}{2})}{sin(x+h)sin\,x}\)]
=\(\lim_{h\rightarrow 0}\) \(\frac{-cos\frac{2x+h}{2}.\frac{sin\frac{h}{2}}{\frac{h}{2}}}{sin(x+h)sin\,x}\)
=(\(\frac{-cos\,x}{sin\,x.\sin\,x}\)).1
=-cosecx cotx ...(3)
From (1), (2), and (3), we obtain
f'(x)=-3cosec²x-5cosec x cotx

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(vi) Let f (x) = 5sinx-6cosx+7. Accordingly, from the first principle,
f'(x) = \(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) [5sin(x+4)-6 cos(x+h)+7-5 sin x+6cosx-7]
= \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) [5{sin(x+h)-sin x}-6(cos(x+h)-cos x}]
=5 \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[sin(x+4)- sin x]-6\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[cos(x+h)-cos.x]
=5\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[2cos(x+h+\(\frac{x}{2}\)) sin(x+h-\(\frac{x}{2}\)) ] -6\(\lim_{h\rightarrow 0}\) cosxcosh – sin x sinh – \(\frac{cos\,x}{h}\)
=5\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[2cos(\(\frac{2x+h}{2}\)) sin(\(\frac{h}{2}\)) ] -6\(\lim_{h\rightarrow 0}\) [\(\frac{-cos\,x(1-cos\,h)-sin\,xsin\,h}{h}\)]
=5 \(\lim_{h\rightarrow 0}\) \({2cos(\frac{2x+h}{2})\frac{sin\frac{h}{2}}{\frac{h}{2}}}-6\lim_{h\rightarrow 0}[-cosx\frac{1-cos\,h}{h}-sin\,x\frac{sin\,h}{h}]\)
= 5 cos x. 1-6(-cos x).(0)-sin.x.1]
=5cosx+6sinx

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(vii) Let f (x) = 2 tan x - 7 sec x. Accordingly, from the first principle,
f'(x) = \(\lim_{h\rightarrow 0}\) \(\frac{f(x+h)-f(x)}{h}\)
= \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[2tan(x+h)-7 sec (x+h)-2 tan x+7 sec x]
= \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[2{tan (x+)-tan x}-7{sec(x+h)-sec.x}]
= 2\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\) [tan (x+h)-tan x]-7 \(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[sec(x+h)-secx]
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[\(\frac{sin(x+h)}{cos(x+h)}-\frac{sin\,x}{cos\,x}\)] -7\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)[\(\frac{1}{cos(x+h)}-\frac{1}{cos\,x}\)]
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)\([\frac{sin(x+h)cos\,x-sin\,xcos(x+h)}{cosxcos(x+h)}]-7\lim_{h\rightarrow 0}\frac{1}{h}[\frac{cos\,x-cos(x+h)}{cos\,xcos(x+h)}]\)
=\(\lim_{h\rightarrow 0}\) \(\frac{1}{h}\)\([\frac{sin(x+h-x)}{cos\,xcos(x+h)}]-7\lim_{h\rightarrow 0}\frac{1}{h}[\frac{-2sin\frac{x+x+h}{2}}{cos\,x-cos(x+h)}]\)
=2.1. \(\frac{1}{cos \,x\, cos\,x}\)- 7.1(\(\frac{sin\,x}{cos\,x\,cos\,x}\))
=2 sec²x-7 sec.x tan.x