How do I do this math problem?

f(x)= 1/x, f(x+h) = 1/(x+h). When h is small, you can take out x. Then you have f(x+n) = 1/x 1/(1+h/x). Expand the second term in series.You get
f(x+h) -1/x ( 1- h/x+h^2/x^2- ...). Now subtract f(x) . You get f(x+h) -f(x) = - h/x^2 + h^3/x^3 - .... Divide it by h and take the limit h tends to 0. You get
lim h tends to 0 will be (f(x+h)-f(x))/h = -1/x^2. This will be the answer.

So if you have:

f(x) = 1/x

Then want to solve for:

[ f(x + h) - f(x) ] / h

Then first we need to get expressions for f(x + h). So substitute (x + h) for x:

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

Now substitute the two expressions above for the two functions, then simplify:

[ 1/(x + h) - 1/x ] / h

Common denominator of x(x + h), so:

{ 1/[x(x + h)] - (x + h)/[x(x + h)]} / h

Now we can subtract the numerators:

{ [1 - (x + h)] / [x(x + h)]} / h

Now simplify the numerator's numerator:

{ (1 - x - h) / [x(x + h)]} / h

Next, turn the division of fractions into the product of the reciprocal and simplify:

{ (1 - x - h) / [x(x + h)]} * 1 / h
(1 - x - h) / [hx(x + h)]

Usually problems like this will allow you to cancel things out, but in this case, this is as simplified as you can get.

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