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Mathematics Question on Derivatives

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

(i) Letf (x) = sin x cos x. Accordingly, from the first principle,
f'(x)= lim⁡h→0\lim_{h\rightarrow 0} f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h}
= lim⁡h→0\lim_{h\rightarrow 0} sin(x+h)cos(x+h)−sin xcos xh\frac{sin(x+h)cos(x+h)-sin\,xcos\,x}{h}
= lim⁡h→0\lim_{h\rightarrow 0} 12h\frac{1}{2h}[sin 2(x+h)-sin 2x]
= lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[ cos4x+2h2\frac{4x+2h}{2} sin 2h2\frac{2h}{2}]
= lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}cos(2x+h) lim⁡h→0\lim_{h\rightarrow 0} sin hh\frac{sin\,h}h{}
= cos(2x+0).1
= cos 2.x

(ii) Letf (x) = sec x. Accordingly, from the first principle,
f'(x)= \lim_{h\rightarrow 0}$$\frac{f(x+h)-f(x)}{h}
= lim⁡h→0\lim_{h\rightarrow 0} sec(x+h)−sec(x)h\frac{sec(x+h)-sec(x)}{h}
= lim⁡h→0\lim_{h\rightarrow 0} 1h[1cos(x+h)−1cos x\frac{1}{cos(x+h)}-\frac{1}{cos\,x}]
= lim⁡h→0\lim_{h\rightarrow 0} 1h1\frac{1}{h_1}[ cos x−cos(x+h)cos xcos(x+h)\frac{cos\,x-cos(x+h)}{cos\,xcos(x+h)}]
=1cos x\frac{1}{cos\,x} lim⁡h→0\lim_{h\rightarrow 0}[-2sin (x+x+h2\frac{h}{2}) sin(x-x-h2\frac{h}{2}) / cos(x+h)]
=1cos x\frac{1}{cos\,x} lim⁡h→0\lim_{h\rightarrow 0}[−2sin(2x+h2)sin(h2)cos(x+h)\frac{-2sin(\frac{2x+h}{2})sin(\frac{h}{2})}{cos(x+h)}]
=\frac{1}{cos\,x}$$\lim_{h\rightarrow 0}[-2sin (2x+h2\frac{2x+h}{2}) sin(h2\frac{h}{2})cos(x+h)]
=1cos x\frac{1}{cos\,x} lim⁡h2→0\lim_{\frac{h}{2}\rightarrow 0} sinh2h2\frac{sin\frac{h}{2}}{\frac{h}{2}}.lim⁡h→0\lim_{h\rightarrow 0} sin(2x+h2)cos(x+h)\frac{sin(\frac{2x+h}{2})}{cos(x+h)}
=1cosx\frac{1}{cosx} .1.sin xcos x\frac{sin\,x}{cos\,x}
= sec x tan x

(iii) Letf (x) = 5 sec x + 4 cos x. Accordingly, from the first principle,
f'(x)= \lim_{h\rightarrow 0}$$\frac{f(x+h)-f(x)}{h}
= lim⁡h→0\lim_{h\rightarrow 0} 5 sec(x+h)+4cos(x+h)−[5 sec x+cos x]h\frac{5\,sec(x+h)+4cos(x+h)-[5\,sec\,x+cos\,x]}{h}
= 5\lim_{h\rightarrow 0}$$\frac{sec(x+h)-sec(x)}{h} +4 \lim_{h\rightarrow 0}$$\frac{cos(x+h)-cos(x)}{h}
= 5lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h} [cos x−cos(x+h)cos xcos(x+h)][\frac{cos\,x-cos(x+h)}{cos\,xcos(x+h)}] +4lim⁡h→0\lim_{h\rightarrow 0} [cosxcosh–sinxsinh −cosx]
=\frac{5}{cosx}$$\lim_{h\rightarrow 0}$$[\frac{sin\frac{2x+h}{2}.sin(\frac{sin(\frac{-h}{2})}{\frac{h}{2}})}{cos(x+h)}]
=5cosx\frac{5}{cosx} [lim⁡h→0\lim_{h\rightarrow 0} sin2x+h2cos(x+h)\frac{sin\frac{2x+h}{2}}{cos(x+h)} . \lim_{h\rightarrow 0}$$\frac{sin(-\frac{h}{2})}{\frac{h}{2}}] - 4sinx
=5cosx\frac{5}{cosx} .sin xcos x\frac{sin\,x}{cos\,x}.1-4sinx
=5sec x tan x-4sin x

(iv) Let f (x) = cosec x. Accordingly, from the first principle,
f'(x) = lim⁡h→0\lim_{h\rightarrow 0} f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h}
f'(x)=lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[cosec(x+h)−cosec xcosec(x+h)-cosec\,x]
=lim⁡h→0\lim_{h\rightarrow 0} \frac{1}{h}$$[\frac{1}{sin(x+h)}-\frac{1}{sin\,x}]
=lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[sin x−sin(x+h)sin(x+h)sin x\frac{sin\,x-sin(x+h)}{sin(x+h)sin\,x}]
=lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h} [2cos(x+x+h2\frac{x+x+h}{2}). sin(x−x−h2)sin(x+h)sin x\frac{sin(\frac{x-x-h}{2})}{sin(x+h)sin\,x}]
=lim⁡h→0\lim_{h\rightarrow 0} [cos2x+h2.sin(h2)sin(x+h)sin x\frac{cos\frac{2x+h}{2}.sin(\frac{h}{2})}{sin(x+h)sin\,x}]
=lim⁡h→0\lim_{h\rightarrow 0} [−cos(2x+h2).(sinh2h2)sin(x+h)sin x][\frac{-cos(\frac{2x+h}{2}).(\frac{sin\frac{h}{2}}{\frac{h}{2}})}{sin(x+h)sin\,x}]
=(−cos xsinx.sinx\frac{-cos\,x}{sinx\\.sinx}).1
=-cosecx cotx

(v) Let f (x) = 3cot x + 5cosec x. Accordingly, from the first principle,
f'(x) = lim⁡h→0\lim_{h\rightarrow 0} f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h}
= lim⁡h→0\lim_{h\rightarrow 0} 3 cot(x+h+5cosec(x+h))−3 cot x−5 cosec xh\frac{3\,cot(x+h+5cosec(x+h))-3\,cot\,x-5\,cosec\,x}{h}
=3lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[cot (x+4)-cot.x]+5lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h} [cosec (x+h) - cosec x] ...(1)
Now, lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h} [cot (x+h)-cot x]
=\lim_{h\rightarrow 0}$$\frac{1}{h} [cos(x+h)sin(x+h)−cos xsin x][\frac{cos(x+h)}{sin(x+h)}-\frac{cos\,x}{sin\,x}]
=lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h} [sin(x−x−h)sin xsin(x+h)\frac{sin(x-x-h)}{sin\,xsin(x+h)}]
=lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[sin(−h)sin x sin(x+4)\frac{sin(-h)}{sin\,x\,sin(x+4)}]
=-(lim⁡h→0\lim_{h\rightarrow 0} sin hh\frac{sin\,h}{h}).(lim⁡h→0\lim_{h\rightarrow 0} 1sin x\frac{1}{sin\,x}-sin(x+h)
=-1. 1sin x\frac{1}{sin\,x} sin(x+0) = −1sinx-\frac{1}{sin^x} = -cosec2x ...(2)
=lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h} [cosec(x+h)-cosecx]
=\lim_{h\rightarrow 0}$$\frac{1}{h}[1sin(x+h)−1sin x\frac{1}{sin(x+h)}-\frac{1}{sin\,x}]
=lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[sin x−sin(x+h)sin(x+h)sin x\frac{sin\,x-sin(x+h)}{sin(x+h)sin\,x}]
=lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h} [2cos(x+x+12).sin(x−x−h2)sin(x+h)sin x\frac{2cos(x + x +\frac{1}{2}). sin (x-x-\frac{h}{2})}{sin(x+h)sin\,x}]
=lim⁡h→0\lim_{h\rightarrow 0} −cos2x+h2.sinh2h2sin(x+h)sin x\frac{-cos\frac{2x+h}{2}.\frac{sin\frac{h}{2}}{\frac{h}{2}}}{sin(x+h)sin\,x}
=(−cos xsin x.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

(vi) Let f (x) = 5sinx-6cosx+7. Accordingly, from the first principle,
f'(x) = lim⁡h→0\lim_{h\rightarrow 0} f(x+h)−f(x)h\frac{f(x+h)-f(x)}{h}
= lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h} [5sin(x+4)-6 cos(x+h)+7-5 sin x+6cosx-7]
= lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h} [5{sin(x+h)-sin x}-6(cos(x+h)-cos x}]
=5 lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[sin(x+4)- sin x]-6lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[cos(x+h)-cos.x]
=5lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[2cos(x+h+x2\frac{x}{2}) sin(x+h-x2\frac{x}{2}) ] -6lim⁡h→0\lim_{h\rightarrow 0} cosxcosh – sin x sinh – cos xh\frac{cos\,x}{h}
=5lim⁡h→0\lim_{h\rightarrow 0} 1h\frac{1}{h}[2cos(2x+h2\frac{2x+h}{2}) sin(h2\frac{h}{2}) ] -6lim⁡h→0\lim_{h\rightarrow 0} [−cos x(1−cos h)−sin xsin hh\frac{-cos\,x(1-cos\,h)-sin\,xsin\,h}{h}]
=5 lim⁡h→0\lim_{h\rightarrow 0} 2cos(2x+h2)sinh2h2−6lim⁡h→0[−cosx1−cos hh−sin xsin hh]{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

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