Slippery slopes... another derivative

Slippery slopes... another derivative

Excerpt: This resource discusses how the derivatives of functions like $(2x+1)^{-1}$ can be obtained by viewing them as translations of functions that we already know how to differentiate. This could precede work on the chain rule.

Problem

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 $-1$.)

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/ePU8jFwR/width/600/height/396/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

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

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1. How can you differentiate $\dfrac{1}{x+1}$?

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

Suggestion for part 1

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

Suggestion for part 2

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

Suggestion for part 3

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?

Solution

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?

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 $-\tfrac{1}{2}$.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/RExfGrf4/width/624/height/407/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

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?