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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(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)

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