1
Question
Find the derivative of the following functions from first principle:
(i) −x
(ii) (−x)−1
(iii) sin(x+1)
(iv) cos(x−π8)
(i) −x
(ii) (−x)−1
(iii) sin(x+1)
(iv) cos(x−π8)
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Solution
(i) Let f(x)=−x
We know that f′(x)=limh→0f(x+h)−f(x)h
⇒f′(x)=limh→0−(x+h)−(−x)h
⇒f′(x)=limh→0−x−h+xh
⇒f′(x)=limh→0−hh
⇒f′(x)=limh→0(−1)
∴f′(x)=−1
(ii) Let f(x)=(−x)−1
We know that f′(x)=limh→0f(x+h)−f(x)h
⇒f′(x)=limh→0(−(x+h))−1+(−x)−1h
⇒f′(x)=limh→0(1−(x+h)−(1−x))h
⇒f′(x)=limh→01x−1x+hh
⇒f′(x)=limh→0x+h−xx(x+h)h
⇒f′(x)=limh→0hhx(x+h)
⇒f′(x)=limh→01x(x+h)
⇒f′(x)=1x(x+0)
∴f′(x)=1x2
(iii) Let f(x)=sin(x+1)
We know that f′(x)=limh→0f(x+h)−f(x)h
⇒f′(x)=limh→0sin((x+h)+1)−sin(x+1)h
⇒f′(x)=limh→0sin(x+1+h)−sin(x+1)h
Using (sinA−sinB=2cos(A+B2)sin(A−B2))
⇒f′(x)=limh→02cos((x+1+h)+(x+1)2)⋅sin((x+1+h)−(x+1)2)h
⇒f′(x)=limh→02cos(2(x+1)+h)2⋅sinh2h
⇒f′(x)=limh→0cos(2(x+1)+h)2⋅sinh2h2
⇒f′(x)=limh→0cos(2(x+1)+h)2⋅limh→0sinh2h2
⇒f′(x)=limh→0cos(2(x+1)+h)2⋅1
⇒f′(x)=cos(2(x+1)+0)2
⇒f′(x)=cos(x+1)
(iv) Let f(x)=cos(x−π8)
We know that f′(x)=limh→0f(x+h)−f(x)h
⇒f′(x)=limh→0cos((x+h)−π8)−cos(x−π8)h
⇒f′(x)=limh→0cos(x−π8+h)−cos(x−π8)h
Using cosA−cosB=−2sin(A+B2)⋅sin(A−B2)
⇒f′(x)=limh→0−2sin⎛⎜
⎜⎝(x−π8+h)+(x−π8)2⎞⎟
⎟⎠⋅sin⎛⎜
⎜⎝(x−π8+h)−(x−π8)2⎞⎟
⎟⎠h
⇒f′(x)=limh→0−2sin⎛⎜
⎜
⎜
⎜⎝2(x−π8+h)2⎞⎟
⎟
⎟
⎟⎠⋅sin(h2)h
⇒f′(x)=−limh→0sin(x−π8+h2)⋅sin(h2)h2
⇒f′(x)=⎡⎢
⎢
⎢⎣limh→0sin(x−π8+h2)×limh→0sinh2h2⎤⎥
⎥
⎥⎦
⇒f′(x)=[sin(x−π8+0)×1]
∴f′(x)=−sin(x−π8)
We know that f′(x)=limh→0f(x+h)−f(x)h
⇒f′(x)=limh→0−(x+h)−(−x)h
⇒f′(x)=limh→0−x−h+xh
⇒f′(x)=limh→0−hh
⇒f′(x)=limh→0(−1)
∴f′(x)=−1
(ii) Let f(x)=(−x)−1
We know that f′(x)=limh→0f(x+h)−f(x)h
⇒f′(x)=limh→0(−(x+h))−1+(−x)−1h
⇒f′(x)=limh→0(1−(x+h)−(1−x))h
⇒f′(x)=limh→01x−1x+hh
⇒f′(x)=limh→0x+h−xx(x+h)h
⇒f′(x)=limh→0hhx(x+h)
⇒f′(x)=limh→01x(x+h)
⇒f′(x)=1x(x+0)
∴f′(x)=1x2
(iii) Let f(x)=sin(x+1)
We know that f′(x)=limh→0f(x+h)−f(x)h
⇒f′(x)=limh→0sin((x+h)+1)−sin(x+1)h
⇒f′(x)=limh→0sin(x+1+h)−sin(x+1)h
Using (sinA−sinB=2cos(A+B2)sin(A−B2))
⇒f′(x)=limh→02cos((x+1+h)+(x+1)2)⋅sin((x+1+h)−(x+1)2)h
⇒f′(x)=limh→02cos(2(x+1)+h)2⋅sinh2h
⇒f′(x)=limh→0cos(2(x+1)+h)2⋅sinh2h2
⇒f′(x)=limh→0cos(2(x+1)+h)2⋅limh→0sinh2h2
⇒f′(x)=limh→0cos(2(x+1)+h)2⋅1
⇒f′(x)=cos(2(x+1)+0)2
⇒f′(x)=cos(x+1)
(iv) Let f(x)=cos(x−π8)
We know that f′(x)=limh→0f(x+h)−f(x)h
⇒f′(x)=limh→0cos((x+h)−π8)−cos(x−π8)h
⇒f′(x)=limh→0cos(x−π8+h)−cos(x−π8)h
Using cosA−cosB=−2sin(A+B2)⋅sin(A−B2)
⇒f′(x)=limh→0−2sin⎛⎜ ⎜⎝(x−π8+h)+(x−π8)2⎞⎟ ⎟⎠⋅sin⎛⎜ ⎜⎝(x−π8+h)−(x−π8)2⎞⎟ ⎟⎠h
⇒f′(x)=limh→0−2sin⎛⎜ ⎜ ⎜ ⎜⎝2(x−π8+h)2⎞⎟ ⎟ ⎟ ⎟⎠⋅sin(h2)h
⇒f′(x)=−limh→0sin(x−π8+h2)⋅sin(h2)h2
⇒f′(x)=⎡⎢ ⎢ ⎢⎣limh→0sin(x−π8+h2)×limh→0sinh2h2⎤⎥ ⎥ ⎥⎦
⇒f′(x)=[sin(x−π8+0)×1]
∴f′(x)=−sin(x−π8)
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