f(x) = 3/2x + 1, with x= a, and x = a + h?

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Method 1: substitute x+h for x.
a)

f(x+h)=2(x+h)+3=2x+2h+3
f(x)=2x+3
[f(x+h)-f(x)]/h=(2x+2h+3-2x-3)/h=2h/h=...


b)

f(x+h)=1/[x+h+1]
f(x)=1/(x+1)

So, we have:
1/[a+b]-1/[a]=
[a]/[a^2+ab]-[a+b]/[a^2+ab]=
[a-a-b]/[a^2+ab]=
[-b]/[a^2+ab]
for a=x+1, b=h,
[-b]/[a^2+ab]=
-h/[x^2+2x+1+xh+h]

then we have:
(a/b)/c=a/(bc)
for a=-c,
-c/bc=(-c/c)(1/b)=
-1/b
for a=-h, c=h, b=x^2+2x+1+xh+h
now the limit of h approaching 0 causes this to become:
b=x^2+2x+1=(x+1)^2
-1/(x+1)^2=
-(x+1)^-2



c)

f(x+h)=(x+h)^2=x^2+2xh+h^2
f(x+h)-f(x)=x^2+2xh+h^2-x^2=2xh+h^2
Divide this by h to get:
2x+h
limit as h approaches 0 makes this:
2x




Method 2: just find the derivative
f(x)=x^a
the derivate:
f'(x)=ax^(a-1)
and if there are more terms we find the derivative of each separate term.


So,

a)

f(x)=2x+3
f(x)=2x^1+3x^0
f'(x)=(2*1)x^(1-1)+(3*0)x^(0-1)
f'(x)=2x^0+0
f'(x)=2

b)
f(x)=1/(x+1)
f(x)=(x+1)^-1
f(x)=1(x+1)^-1
f'(x)=(1*-1)(x+1)^(-1-1)
f'(x)=-1(x+1)^-2
f'(x)=-(x+1)^-2

c)
f(x)=x^2
f(x)=1x^2
f'(x)=(1*2)x^(2-1)
f'(x)=2x^1
f'(x)=2x

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