You might find some of the ideas in Slippery slopes helpful when thinking about this problem.

Here are the graphs of \(y=\dfrac{1}{x}\) and \(y=\dfrac{1}{x+1}\) with tangents drawn at points A and B.

What happens when you move the point A? (Be careful when the \(x\)-coordinate of A gets close to \(-1\).)

  1. How could you work out the gradient of \(y=\dfrac{1}{x+1}\) at the point \((2,\tfrac{1}{3})\)?

  2. How can you differentiate \(\dfrac{1}{x+1}\)?

  3. How could you adapt these approaches to work out the derivative of \(\dfrac{1}{2x+1}\)?

The graph of \(y=\dfrac{1}{x+1}\) is a translation of the graph of \(y=\dfrac{1}{x}\).

Can you generalise ideas from part 1 to find the derivative of \(\dfrac{1}{x+1}\) at a general point \((x,y)\)?

We already know how to differentiate constant multiples of powers of \(x\), so we know how to differentiate \(\dfrac{1}{2x}\).

Can you suggest other examples where you could use a similar approach to find derivatives?

In this problem we have only composed the function \(\dfrac{1}{x}\) with functions of the form \(ax+b\). Could you apply any of these ideas to polynomial, rational or trigonometric functions?