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 \(-1\).)*

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

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

There are several ways to find the gradient function or derivative of \(\dfrac{1}{x+1}.\) The approaches we discuss here assume that you do not already know how to differentiate this function directly.

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

Therefore the tangent to \(y=\dfrac{1}{x+1}\) at \((2,\tfrac{1}{3})\) is parallel to the tangent to \(y=\dfrac{1}{x}\) at \((3, \tfrac{1}{3}).\)

It follows that the gradient of \(y=\dfrac{1}{x+1}\) where \(x=2\) is the same as the gradient of \(\dfrac{1}{x}\) where \(x=3.\)

Therefore the gradient of \(y=\dfrac{1}{x+1}\) at \((2,\tfrac{1}{3})\) is \(-\tfrac{1}{9}.\)

Now consider a general point \((x,y)\) on the graph of \(y=\dfrac{1}{x+1}.\)

From the ideas in part 1, the gradient of \(y=\dfrac{1}{x+1}\) at a general point \((x,y)\) is the same as the gradient of \(y=\dfrac{1}{x}\) at the point with coordinates \((x+1,y)\).

You can see this by moving the point A in the GeoGebra file above.

Therefore the derivative of \(\dfrac{1}{x+1}\) is \(\dfrac{-1}{(x+1)^{2}}.\)

What would happen if the function we wanted to differentiate had been a translation of \(y=\dfrac{1}{x}\) parallel to the \(y\)-axis instead?

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

We already know how to differentiate constant multiples of powers of \(x\), so we know how to differentiate \(\dfrac{1}{2x}\). From the first part of the problem we also know a way of finding the derivatives of horizontal translations of functions. What is \(y=\dfrac{1}{2x+1}\) as a translation of \(y=\dfrac{1}{2x}\)?

The vertical asymptotes of the two graphs suggest that \(y=\dfrac{1}{2x+1}\) is a translation of \(y=\dfrac{1}{2x}\) by \(\dfrac{1}{2}\) in the negative \(x\)-direction.

Function notation can help to show this too.

If we write \(\dfrac{1}{2x}=f(x)\) then \(\dfrac{1}{2x+1} = \dfrac{1}{2(x+\tfrac{1}{2})} = f(x+\tfrac{1}{2}).\)

This means that the gradient of \(y=\dfrac{1}{2x+1}\) at \((x,y)\) is the same as the gradient of \(y=\dfrac{1}{2x}\) at \((x+\tfrac{1}{2}, y)\).

You can see this by moving the point A in the GeoGebra file below. *Be careful when the \(x\)-coordinate of A gets close to \(-\tfrac{1}{2}\).*

It follows that the gradient of \(y=\dfrac{1}{2x+1}\) at \((x,y)\) is \(\dfrac{-1}{2(x+\tfrac{1}{2})^{2}}.\)

We can write this as \(\dfrac{-2}{(2x+1)^{2}}\) to help see the relationship between the original function and its derivative.

You could also think of \(y=\dfrac{1}{2x+1}\) as a transformation of \(y=\dfrac{1}{x}.\) How would this affect your approach?