Okay, let's differentiate these functions using the first principle.
The definition of the derivative from first principles is:
f′(x)=limh→0hf(x+h)−f(x)
iii) y=4x2+2x+1
Step 1: Identify f(x) and f(x+h).
f(x)=4x2+2x+1
f(x+h)=4(x+h)2+2(x+h)+1
f(x+h)=4(x2+2xh+h2)+2x+2h+1
f(x+h)=4x2+8xh+4h2+2x+2h+1
Step 2: Calculate the difference f(x+h)−f(x).
f(x+h)−f(x)=(4x2+8xh+4h2+2x+2h+1)−(4x2+2x+1)
f(x+h)−f(x)=8xh+4h2+2h
Step 3: Substitute into the limit definition and simplify.
f′(x)=limh→0h8xh+4h2+2h
f′(x)=limh→0hh(8x+4h+2)
f′(x)=limh→0(8x+4h+2)
Step 4: Evaluate the limit.
f′(x)=8x+4(0)+2
f′(x)=8x+2
iv) y=sinx
Step 1: Identify f(x) and f(x+h).
f(x)=sinx
f(x+h)=sin(x+h)
Step 2: Calculate the difference f(x+h)−f(x).
f(x+h)−f(x)=sin(x+h)−sinx
Using the sum-to-product identity sinA−sinB=2cos(2A+B)sin(2A−B):
f(x+h)−f(x)=2cos(2x+h+x)sin(2x+h−x)
f(x+h)−f(x)=2cos(x+2h)sin(2h)
Step 3: Substitute into the limit definition and simplify.
f′(x)=limh→0h2cos(x+2h)sin(2h)
f′(x)=limh→0[cos(x+2h)⋅2hsin(2h)]
Step 4: Evaluate the limit using limθ→0θsinθ=1.
f′(x)=cos(x+0)⋅1
f′(x)=cosx
v) y=cosx
Step 1: Identify f(x) and f(x+h).
f(x)=cosx
f(x+h)=cos(x+h)
Step 2: Calculate the difference f(x+h)−f(x).
f(x+h)−f(x)=cos(x+h)−cosx
Using the sum-to-product identity cosA−cosB=−2sin(2A+B)sin(2A−B):
f(x+h)−f(x)=−2sin(2x+h+x)sin(2x+h−x)
f(x+h)−f(x)=−2sin(x+2h)sin(2h)
Step 3: Substitute into the limit definition and simplify.
f′(x)=limh→0h−2sin(x+2h)sin(2h)
f′(x)=limh→0[−sin(x+2h)⋅2hsin(2h)]
Step 4: Evaluate the limit using limθ→0θsinθ=1.
f′(x)=−sin(x+0)⋅1
f′(x)=−sinx
vi) y=tanx
Step 1: Identify f(x) and f(x+h).
f(x)=tanx=cosxsinx
f(x+h)=tan(x+h)=cos(x+h)sin(x+h)
Step 2: Calculate the difference f(x+h)−f(x).
f(x+h)−f(x)=cos(x+h)sin(x+h)−cosxsinx
f(x+h)−f(x)=cos(x+h)cosxsin(x+h)cosx−cos(x+h)sinx
Using the angle subtraction formula sin(A−B)=sinAcosB−cosAsinB:
f(x+h)−f(x)=cos(x+h)cosxsin((x+h)−x)=cos(x+h)cosxsinh
Step 3: Substitute into the limit definition and simplify.
f′(x)=limh→0hcos(x+h)cosxsinh
f′(x)=limh→0[hsinh⋅cos(x+h)cosx1]
Step 4: Evaluate the limit using limθ→0θsinθ=1.
f′(x)=1⋅cos(x+0)cosx1
f′(x)=cosxcosx1=cos2x1