def of derivative
f'(x)=limh->0(f(x+h)-f(x))/(h)
Substitution
f'(x)=limh->0(2/(x+h+4)-2/(x+4))/(h)
Common Denominator
f'(x)=limh->0((2(x+4))/((x+4)(x+h+4))-(2(x+h+4))/((x+4)(x+h+4)))/(h)
Distribute and write as a single numerator
f'(x)=limh->0((2x+8)/((x+4)(x+h+4))-(2x+2h+8)/((x+4)(x+h+4)))/(h)
f'(x)=limh->0((2x+8-2x-2h-8)/((x+4)(x+h+4)))/(h)
Simplify
f'(x)=limh->0((cancel(2x)cancel(+8)cancel(-2x)-2hcancel(-8))/((x+4)(x+h+4)))/(h)
f'(x)=limh->0((-2h)/((x+4)(x+h+4)))/(h)
Multiply by the reciprocal
f'(x)=limh->0(-2h)/((x+4)(x+h+4))*(1/h)
f'(x)=limh->0(-2h)/(h(x+4)(x+h+4))
Simplify
f'(x)=limh->0(-2cancelh)/(cancelh(x+4)(x+h+4))
f'(x)=limh->0(-2)/((x+4)(x+h+4))
Now we can substitute in a 0 for h
f'(x)=(-2)/((x+4)(x+0+4))
Simplify
f'(x)=(-2)/((x+4)(x+4))
Simplify
f'(x)=(-2)/((x+4)^2)
Watch this tutorial to see a similar question solved used the same methods.
