Alright IT — let's do this.
To differentiate from first principles, we use the definition of the derivative:
dxdy=limh→0hf(x+h)−f(x)
i) y=x2
Step 1: Identify f(x) and f(x+h).
f(x)=x2
f(x+h)=(x+h)2=x2+2xh+h2
Step 2: Substitute into the limit definition.
dxdy=limh→0h(x2+2xh+h2)−x2
Step 3: Simplify the numerator.
dxdy=limh→0h2xh+h2
Step 4: Factor out h from the numerator and cancel.
dxdy=limh→0hh(2x+h)
dxdy=limh→0(2x+h)
Step 5: Evaluate the limit as h→0.
dxdy=2x+0
dxdy=2x
ii) y=3x+2
Step 1: Identify f(x) and f(x+h).
f(x)=3x+2
f(x+h)=3(x+h)+2=3x+3h+2
Step 2: Substitute into the limit definition.
dxdy=limh→0h(3x+3h+2)−(3x+2)
Step 3: Simplify the numerator.
dxdy=limh→0h3x+3h+2−3x−2
dxdy=limh→0h3h
Step 4: Cancel h.
dxdy=limh→03
Step 5: Evaluate the limit.
dxdy=3
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: Substitute into the limit definition.
dxdy=limh→0h(4x2+8xh+4h2+2x+2h+1)−(4x2+2x+1)
Step 3: Simplify the numerator.
dxdy=limh→0h4x2+8xh+4h2+2x+2h+1−4x2−2x−1
dxdy=limh→0h8xh+4h2+2h
Step 4: Factor out h from the numerator and cancel.
dxdy=limh→0hh(8x+4h+2)
dxdy=limh→0(8x+4h+2)
Step 5: Evaluate the limit as h→0.
dxdy=8x+4(0)+2
dxdy=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: Substitute into the limit definition.
dxdy=limh→0hsin(x+h)−sinx
Step 3: Use the trigonometric identity sinA−sinB=2cos(2A+B)sin(2A−B).
Here A=x+h and B=x.
2A+B=2x+h+x=22x+h=x+2h
2A−B=2x+h−x=2h
So, sin(x+h)−sinx=2cos(x+2h)sin(2h).
Step 4: Substitute back into the limit.
dxdy=limh→0h2cos(x+2h)sin(2h)
Step 5: Rearrange the terms to use the standard limit limθ→0θsinθ=1.
dxdy=limh→0[cos(x+2h)⋅2hsin(2h)]
Step 6: Evaluate the limit. As h→0, 2h→0.
dxdy=cos(x+0)⋅1
dxdy=cosx
v) y=cosx
Step 1: Identify f(x) and f(x+h).
f(x)=cosx
f(x+h)=cos(x+h)
Step 2: Substitute into the limit definition.
dxdy=limh→0hcos(x+h)−cosx
Step 3: Use the trigonometric identity cosA−cosB=−2sin(2A+B)sin(2A−B).
Here A=x+h and B=x.
2A+B=x+2h
2A−B=2h
So, cos(x+h)−cosx=−2sin(x+2h)sin(2h).
Step 4: Substitute back into the limit.
dxdy=limh→0h−2sin(x+2h)sin(2h)
Step 5: Rearrange the terms to use the standard limit limθ→0θsinθ=1.
dxdy=limh→0[−sin(x+2h)⋅2hsin(2h)]
Step 6: Evaluate the limit. As h→0, 2h→0.
dxdy=−sin(x+0)⋅1
dxdy=−sinx
vi) y=tanx
Step 1: Identify f(x) and f(x+h).
f(x)=tanx
f(x+h)=tan(x+h)
Step 2: Substitute into the limit definition.
dxdy=limh→0htan(x+h)−tanx
Step 3: Rewrite tanx=cosxsinx.
dxdy=limh→0hcos(x+h)sin(x+h)−cosxsinx
Step 4: Combine the fractions in the numerator.
dxdy=limh→0hcos(x+h)cosxsin(x+h)cosx−cos(x+h)sinx
Step 5: Use the trigonometric identity sinAcosB−cosAsinB=sin(A−B).
Here A=x+h and B=x. So, sin((x+h)−x)=sinh.
dxdy=limh→0hcos(x+h)cosxsinh
Step 6: Separate the limit using limh→0hsinh=1.
dxdy=limh→0(hsinh⋅cos(x+h)cosx1)
Step 7: Evaluate the limit.
dxdy=1⋅cos(x+0)cosx1
dxdy=cosxcosx1=cos2x1
dxdy=sec2x
vii) y=x3+x2−x+4
Step 1: Identify f(x) and f(x+h).
f(x)=x3+x2−x+4
f(x+h)=(x+h)3+(x+h)2−(x+h)+4
Expand the terms:
(x+h)3=x3+3x2h+3xh2+h3
(x+h)2=x2+2xh+h2
So,
f(x+h)=(x3+3x2h+3xh2+h3)+(x2+2xh+h2)−(x+h)+4
f(x+h)=x3+3x2h+3xh2+h3+x2+2xh+h2−x−h+4
Step 2: Substitute into the limit definition.
dxdy=limh→0h(x3+3x2h+3xh2+h3+x2+2xh+h2−x−h+4)−(x3+x2−x+4)
Step 3: Simplify the numerator by canceling terms.
dxdy=limh→0h3x2h+3xh2+h3+2xh+h2−h
Step 4: Factor out h from the numerator and cancel.
dxdy=limh→0hh(3x2+3xh+h2+2x+h−1)
