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Question

Find the derivative of the following functions from first principle:
(i) x
(ii) (x)1
(iii) sin(x+1)
(iv) cos(xπ8)

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Solution

(i) Let f(x)=x
We know that f(x)=limh0f(x+h)f(x)h
f(x)=limh0(x+h)(x)h
f(x)=limh0xh+xh
f(x)=limh0hh
f(x)=limh0(1)
f(x)=1

(ii) Let f(x)=(x)1
We know that f(x)=limh0f(x+h)f(x)h
f(x)=limh0((x+h))1+(x)1h
f(x)=limh0(1(x+h)(1x))h
f(x)=limh01x1x+hh
f(x)=limh0x+hxx(x+h)h
f(x)=limh0hhx(x+h)
f(x)=limh01x(x+h)
f(x)=1x(x+0)
f(x)=1x2

(iii) Let f(x)=sin(x+1)
We know that f(x)=limh0f(x+h)f(x)h
f(x)=limh0sin((x+h)+1)sin(x+1)h
f(x)=limh0sin(x+1+h)sin(x+1)h

Using (sinAsinB=2cos(A+B2)sin(AB2))
f(x)=limh02cos((x+1+h)+(x+1)2)sin((x+1+h)(x+1)2)h
f(x)=limh02cos(2(x+1)+h)2sinh2h
f(x)=limh0cos(2(x+1)+h)2sinh2h2
f(x)=limh0cos(2(x+1)+h)2limh0sinh2h2
f(x)=limh0cos(2(x+1)+h)21
f(x)=cos(2(x+1)+0)2
f(x)=cos(x+1)

(iv) Let f(x)=cos(xπ8)
We know that f(x)=limh0f(x+h)f(x)h
f(x)=limh0cos((x+h)π8)cos(xπ8)h
f(x)=limh0cos(xπ8+h)cos(xπ8)h

Using cosAcosB=2sin(A+B2)sin(AB2)

f(x)=limh02sin⎜ ⎜(xπ8+h)+(xπ8)2⎟ ⎟sin⎜ ⎜(xπ8+h)(xπ8)2⎟ ⎟h

f(x)=limh02sin⎜ ⎜ ⎜ ⎜2(xπ8+h)2⎟ ⎟ ⎟ ⎟sin(h2)h

f(x)=limh0sin(xπ8+h2)sin(h2)h2
f(x)=⎢ ⎢ ⎢limh0sin(xπ8+h2)×limh0sinh2h2⎥ ⎥ ⎥
f(x)=[sin(xπ8+0)×1]
f(x)=sin(xπ8)

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