Building blocks

## Solution

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 $\dfrac{1}{x+1}$?

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?

1. 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?