Well, you could break it apart (I think...), or do it all at once. I'll do the latter.
[f(x + h) - f(x)] / h
[((2(x + h) + 1) / (x + h) - (2x+1) / x] / h
[(2x + 2h + 1)/(x + h) - (2x + 1)/x] / h
Now comes the "fun" part, I'd combine what I have in the [] to one fraction by multiplying one fraction by the denominator of the opposite and dividing by the product of denominators. Also, Since you're dividing everything by h, just throw h in the denominator as well
[x(2x + 2h + 1) - (2x + 1)(x + h)]/[hx(x + h)]
Simplify the numerator by factoring...
[2x² + 2hx + x - 2x² - 2xh - x - h]/[hx(x + h)]
[(2x² - 2x²) + (2hx - 2hx) + (x - x) - h]/[hx(x + h)]
-h/[hx(x + h)]
-1/[x(x + h)]
Thus, the difference quotient is -1/[x(x + h)].
Note that as a check, when h → 0, you should get the derivative (which you probably don't know yet if you're learning this now). So the derivative should be -1/x². As a check, http://www.wolframalpha.com/input/?i=der…
[f(x + h) - f(x)] / h
[((2(x + h) + 1) / (x + h) - (2x+1) / x] / h
[(2x + 2h + 1)/(x + h) - (2x + 1)/x] / h
Now comes the "fun" part, I'd combine what I have in the [] to one fraction by multiplying one fraction by the denominator of the opposite and dividing by the product of denominators. Also, Since you're dividing everything by h, just throw h in the denominator as well
[x(2x + 2h + 1) - (2x + 1)(x + h)]/[hx(x + h)]
Simplify the numerator by factoring...
[2x² + 2hx + x - 2x² - 2xh - x - h]/[hx(x + h)]
[(2x² - 2x²) + (2hx - 2hx) + (x - x) - h]/[hx(x + h)]
-h/[hx(x + h)]
-1/[x(x + h)]
Thus, the difference quotient is -1/[x(x + h)].
Note that as a check, when h → 0, you should get the derivative (which you probably don't know yet if you're learning this now). So the derivative should be -1/x². As a check, http://www.wolframalpha.com/input/?i=der…
-
f(x)= 2x + 1
f(x + h) = 2(x + h) + 1
f(x + h) - f(x) = 2(x + h) + 1 - 2x + 1
f(x + h) - f(x) = 2x + 2h - 2x + 2
f(x + h) - f(x) = 2h + 2
[f(x + h) - f(x)] / h = (2h + 2) / h
[f(x + h) - f(x)] / h = (2h / h) + (2 / h)
[f(x +h) - f(x)] / h = 2 + 2/h
[f(x + h) - f(x) ] / h = 2(1 + 1/h)
¯¯¯¯¯¯¯¯…
f(x + h) = 2(x + h) + 1
f(x + h) - f(x) = 2(x + h) + 1 - 2x + 1
f(x + h) - f(x) = 2x + 2h - 2x + 2
f(x + h) - f(x) = 2h + 2
[f(x + h) - f(x)] / h = (2h + 2) / h
[f(x + h) - f(x)] / h = (2h / h) + (2 / h)
[f(x +h) - f(x)] / h = 2 + 2/h
[f(x + h) - f(x) ] / h = 2(1 + 1/h)
¯¯¯¯¯¯¯¯…
-
f(x)= (2x+1)/x
f(x + h) = (2(x + h) + 1)/(x+h)=(2x+1+2h)/(x+h)
f(x + h) - f(x) = ( (2x+1+2h)x - (2x+1)(x+h) ) / ( x(x+h) )
f(x + h) - f(x) = ( 2x^2+x+2hx - (2x^2+x +2xh +h ) / ( x(x+h) )
f(x + h) - f(x) = - h / ( x(x+h) )
hence: (f(x + h) - f(x)) / h = - 1 / ( x(x+h) )
f(x + h) = (2(x + h) + 1)/(x+h)=(2x+1+2h)/(x+h)
f(x + h) - f(x) = ( (2x+1+2h)x - (2x+1)(x+h) ) / ( x(x+h) )
f(x + h) - f(x) = ( 2x^2+x+2hx - (2x^2+x +2xh +h ) / ( x(x+h) )
f(x + h) - f(x) = - h / ( x(x+h) )
hence: (f(x + h) - f(x)) / h = - 1 / ( x(x+h) )