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Question

Find the derivative of functions cosx, cosecx, secx, cotx using first principle​​​​​​

A
sinx, cosecxcotx, secxtanx, cosec2x
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B
sinx, cosecxcotx, secxtanx, cosec2x
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C
sinx, cosecxcotx, secxtanx, cosec2x
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D
sinx, cosecxcotx, secxtanx, cosec2x
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Solution

The correct option is A sinx, cosecxcotx, secxtanx, cosec2x
(i) f(x)=limh0f(x+h)f(x)h
=limh0cos(x+h)cosxh
=limh0)cosxcoshsinxsinhcosxh
=limh0cosx(cosh1)sinxsinhh
=limh0cosx(cosh1)hlimh0sinxsinhh
=cosxlimh0cosh1hsinxlimh0sinhh
=cosx×0sinx×1=sinx

(ii) f(x)=limh0f(x+h)f(x)h
=limh0cosec(x+h)cosecxh
=limh01sin(x+h)1sinxh
=limh0sinxsin(x+h)sin(x+h)sinxh
Using, sinAsinB=2cos(A+B)2sin(AB)2
=limh02cos(x+x+h)2sin(xxh)2hsin(x+h)sinx
=limh02cos(2x+h)2sin(h)2hsin(x+h)sinx
=limh0cos(2x+h)2sin(x+h)sinxlimh0sin(h/2)h2
Second limit has value unity, which is already derived in theory class.
=cos(x)(sin(x)sinx)(1)=cosecxcotx

(iii) f(x)=limh0f(x+h)f(x)h
=limh0sec(x+h)secxh
=limh01cos(x+h)1cosxh
=limh0((cosxcos(x+h))/(cos(x+h)cosx))/h
Using, cosAcosB=2sin(A+B)2sin(AB)2
=limh02sin(x+x+h)2sin(xxh)2hcos(x+h)cosx
=limh02sin(2x+h)2sin(h)2hcos(x+h)cosx
=limh0sin(2x+h)2cos(x+h)cosxlimh0sin(h/2)h2
Second limit has value unity, which is already derived in theory class.
=limh0sin(2x+h)2cos(x+h)cosxlimh0sin(h/2)h2=sin(x)(cos(x)cosx)(1)=secxtanx

(iv) f(x)=limh0f(x+h)f(x)h
=limh0cot(x+h)cotxh
=limh0cos(x+h)sin(x+h)cosxsinxh
=limh0cos(x+h)sinxsin(x+h)cosxsin(x+h)sinxh
Using, sin(AB)=sinAcosBcosAsinB
=limh0sin(xxh)hsin(x+h)sinx
=limh0sin(h)hsin(x+h)sinx
=limh01sin(x+h)sinxlimh0sin(h)h
Second limit has value unity, which is already derived in theory class.
=1sinxsinx(1)=cosec2x


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