I can see $u$!

I can see $u$!

Excerpt: Students are asked to identify a suitable $u$ or ‘inner function’ for each of the given functions and then find $\tfrac{dy}{dx}.$ Different ways of rewriting the expressions involved may lead to alternative options for $u.$

Problem

If $u=2x+1$ and $y=u^3$ what is $y$ as a function $x$?

Working backwards, if $y=(2x+3)^{-2}$, how could you express $y$ as a function of $u$ where $u$ is a function of $x$?

If you also wanted to work out $\dfrac{dy}{dx}$, how would this affect your choice of $u$?

Take a look at the functions in the table below.

Write $y$ as a function of $u$ where $u$ is a function of $x$ so that you can use these to find $\dfrac{dy}{dx}.$

You may be able to do this in more than one way.

Solution

If $u=2x+1$ then substituting $2x+1$ for $u$ in $y=u^3$ gives $y=(2x+1)^3.$

Looking at $y=(2x+3)^{-2}$, I might describe it as “one over something squared”, so one option is to let the “something” be $u.$ In a function machine for $y$, the squaring could happen before or after taking the reciprocal, giving further options.

Here are some possibilities for writing $y$ as a function of $u$ where $u$ is a function of $x.$

• $y=u^{-2}$ and $u=2x+3$

• $y=u^{-1}$ and $u=(2x+3)^2$

• $y=u^2$ and $u=(2x+3)^{-1}$

However, there is no reason why $y$ can’t be the identity function of $u$ or $u$ can’t be the identity function of $x$, so there are two other possibilities.

• $y=(2u+3)^{-2}$ and $u=x$

• $y=u$ and $u=(2x+3)^{-2}$

If you also wanted to work out $\dfrac{dy}{dx}$, how would this affect your choices of $u$?

As $y$ is a function of a function, I can use the chain rule to differentiate it if I can find $\dfrac{dy}{du}$ and $\dfrac{du}{dx}.$ Can I do this for all the possibilities I found for $u$?

For the first case I can find $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$ directly. In the second and third cases, I can find $\dfrac{dy}{du}$ but need to think a bit more about how I could differentiate $u.$ The chain rule could help, or I could look at Slippery slopes …. another derivative to help me. differentiate $(2x+3)^{-1}.$

The final two cases don’t help me either because I’d need the chain rule to find $\dfrac{dy}{du}$ or $\dfrac{du}{dx}.$

I can tackle these functions in any order, but scanning the table suggests a few ways to group the functions. As in the previous example, I could use the identity function and say either $y=u$ or $u=x$, but this won’t necessarily help me to find $\dfrac{dy}{dx}$, so I’ll need to look for other options in most cases. I’ll find possibilities for $u$ first then try to differentiate. There is a table with suggestions for $u$ and expressions for $\dfrac{dy}{dx}$ at the end of this section.

Possibilities for $u$

Powers of a linear polynomial

The functions in the top row are all powers of a linear polynomial. This makes me wonder if I could also express $9x^2-6x+1$ and $\dfrac{1}{x^2+4x+4}$ as powers of a linear polynomial.

For $y=\dfrac{1}{x^2+4x+4}$ I can choose $u=x+2$ and $y=u^{-2}.$ But I could also choose $u=(x+2)^2$ and $y=\dfrac{1}{u}.$ If I now look at $\dfrac{5}{\sqrt{x}}$ with this in mind, I notice that it can also be written as $\sqrt{\dfrac{25}{x}}.$ How does this affect my choices for $u$?

Exponential and logarithmic functions

Several functions in the table involve either exponential or logarithmic functions and I can see more than one $u$ for several of these functions. For example, $e^{5x}$ could be thought of as $5x$ composed with $e^x$, but also as $e^{x}$ composed with $x^5.$ In a similar way, rewriting the logarithms may help me to see different possibilities for $u.$ However, if I am rewriting expressions, I need to think about which values of $x$ the original expression is defined for, and whether my new expression is also defined for all these values of $x.$ For example, where are $\ln{x^2}$ and $2\ln{x}$ defined?

The logarithms in $y=\ln{x^2}+\ln{8x}$ can be combined and separated in different ways, including $\ln{8x^3}$, $\ln{(2x)^3}$ and $3\ln{2x}$, which gives me lots of options for $u.$

Trigonometric functions

Looking now at the trigonometric functions at the bottom left of the table, I can see that $\sin x^2$ is a function of $x^2.$ But how can $y=\cot x$ be treated as a composition of functions? I recall that $\cot x=\dfrac{1}{\tan x}$, which gives me an idea for $u.$

The other functions look a bit more complicated, and I may need to rewrite them using identities. For example, $\sec^2 x=1+\tan^2 x$ so maybe I can express $\tan x (\sec^2 x-1)$ in terms of $\tan x.$

I can think of $\tfrac{1}{2}(1-\cos 2x)$ as a function of $\cos 2x$, or of $2x$, but I also recognise it from the double angle formulae, where $\tfrac{1}{2}(1-\cos 2x)=\sin^2 x.$ This looks like a very different function!

The identity function

The only remaining function in the table is $y=x$, which looks like the simplest function in the table. I could have included $y=x$ in the powers of linear polynomials group, but there may be some other ways of thinking about it.

As a first step, I can think about the function that has no effect on its inputs: the identity function. So I can let $u=x$ and then I need $y=u.$

Looking around the table gives me an idea. I need the composition of two functions to have no effect, but each of the functions can change its inputs into something else. For example, I could let $u=e^x$ and have $y=\ln{u}.$ Similarly, provided $x\neq 0$, I could let $u=\dfrac{1}{x}$ and $y=\dfrac{1}{u}.$ I need to be careful about the domains, but what other functions could I choose?

Looking back, I realise that almost all of the functions in the table can be thought of as a composition of a linear polynomial and another function. It may be interesting to think about how this affects their derivatives.

I also notice that I could think of a few of these functions as a composition of three functions. For example, for $y=\dfrac{1}{x^2+4x+4}$ I could have had $u=x+2$, $v=u^2$ and $y=\dfrac{1}{v}.$ Which others could I break down using three functions instead of two? How do I differentiate a composition of three functions?

Finding the derivatives

Also work out $\dfrac{dy}{dx}$ in each case. How does your choice of $u$ affect how you find $\dfrac{dy}{dx}$?

I can use the chain rule to differentiate these functions. But as I found earlier, some of the possibilities for $u$ may not be as helpful as others. Do I know how to find $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$ for all the possibilities I found for $u$?

Derivatives of the powers of linear polynomials

For the powers of linear polynomials, I’ll just consider a few examples.

For $y=(5-2x)^3$, $u$ could be $5-2x$ and then $y=u^3.$ Applying the chain rule gives $\dfrac{dy}{dx}=3u^2 \cdot -2=-6(5-2x)^2$, so the $-2$ from $5-2x$ is acting as a scale factor in the derivative. If I had $(ax+b)^3$ instead, what could I choose as $a$ and $b$ if I didn’t want to introduce a scale factor in the derivative like this?

Earlier I found that I could write $y=9x^2-6x+1$ as a composition of functions, but I already know how to differentiate it directly, so I could just take $y=u$ and $u=9x^2-6x+1$. How does differentiating $y=9x^2-6x+1$ using these different functions compare with differentiating it directly?

Again, I can differentiate $y=\dfrac{5}{\sqrt{x}}$ directly, so I could have taken $y=u$ and $u=\dfrac{5}{\sqrt{x}}.$ If instead I differentiate $y$ using $u=\sqrt{x}$ and $y=\dfrac{5}{u}$ then I get $\dfrac{dy}{dx}=\dfrac{-5}{u^2}\cdot \dfrac{1}{2\sqrt{x}}=\dfrac{-5}{2x^{3/2}}$

but what happens if I use $u=\dfrac{25}{x}$ and $y=\sqrt{u}$ instead?

Derivatives of the exponential and logarithmic functions

I saw two different possibilities for $u$ for $e^{5x}$ and $e^{x+4}$ and one might make the differentiation simpler than the other.

I also noticed that $e^{5x}$ and $e^{x+4}$ are both functions of linear polynomials, so when I differentiate them using the chain rule will I see the same effect of scale factors in the derivatives as I found for powers of linear polynomials?

Differentiating $y=\ln{3x}$ using $u=3x$ and $y=\ln{u}$ gives me $\dfrac{dy}{dx}=\dfrac{1}{u}\cdot 3=\dfrac{1}{x}$ which seems surprising. I might have expected a scale factor of $3$ in the derivative, but looking at my other option for $u$ helps to confirm that my derivative is correct. Another thing to think about is that $y=\ln{3x}$ is the inverse of an exponential function and I have already differentiated some exponential functions. What is the inverse of $y=\ln{3x}$ and how could I use its derivative to find the derivative of $y=\ln{3x}$?

Looking now at $y=\ln{x^2}$, I can use $u=x^2$ and $y=\ln{u}$ to give $\dfrac{dy}{dx}=\dfrac{1}{u}\cdot 2x=\dfrac{2}{x}$ but what if I had rewritten $y$ as $y=2\ln{x}$ instead?

Turning now to $y=\ln{x^2}+\ln{8x}$, I have lots of options for $u$ based on different ways of rewriting $\ln{x^2}+\ln{8x}$ and some may help me to differentiate the function more simply than others. However, having differentiated $y=\ln{x^2}$ and $y=\ln{3x}$, I can already conjecture what the derivative of $y=\ln{x^2}+\ln{8x}$ is.

Derivatives of the trigonometric functions

Again, I’ll just consider a couple of examples.

Using $u=\tan x$ and $y=\dfrac{1}{u}$ to differentiate $y=\cot x$ gives $\dfrac{dy}{dx}=\dfrac{-1}{u^2}\cdot \sec^{2}x=\dfrac{-\sec^2 x}{\tan^2 x}$ and writing this in terms of $\sin x$ and $\cos x$ could help show how this could be simplified.

I had lots of options for $u$ for $y=\tfrac{1}{2}(1-\cos 2x)$, but the first was $u=\cos 2x$, which I could differentiate using the chain rule or ideas from Sine stretching. Alternatively I could use $u=2x$ and $y=\tfrac{1}{2}-\tfrac{1}{2}\cos u$ since then I can can find $\dfrac{dy}{du}$ directly. Both of these lead me to $\dfrac{dy}{dx}=\sin 2x.$

But I also noticed that I can write $\tfrac{1}{2}(1-\cos 2x)$ as $\sin^2 x.$ How does using $u=\sin x$ and $y=u^2$ lead to the same derivative as the other routes?

Derivative of the identity function

I already know the derivative, but what happens if I write $y=u$ and $u=x$ and use the chain rule?

How do the different possibilities for writing $y=x$ as a composition of functions lead to this same result?

What other expressions could be seen as a composition of functions? Can you suggest some that you can also differentiate? Are there different options for $u$ as there were with many of the functions in this table?