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?
