Fluency exercise

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Solution

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

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

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

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.

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

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.

 $y=(5-2x)^3$ $y=\sqrt{3x-1}$ $y=\dfrac{5}{\sqrt{x}}$ $y=9x^2-6x+1$ $y=e^{5x}$ $y=\ln{3x}$ $y=\dfrac{1}{x^2+4x+4}$ $y=e^{x+4}$ $y=\ln{x^2}$ $y=\cot x$ $y=x$ $y=\ln{x^2}+\ln{8x}$ $y=\sin x^2$ $y=\tan x (\sec^2 x-1)$ $y=\tfrac{1}{2}(1-\cos 2x)$

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 $u$

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

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.

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?

 $y=(5-2x)^3$ $y=\sqrt{3x-1}$ $y=\dfrac{5}{\sqrt{x}}$ $u=5-2x$ and $y=u^3$ $u=3x-1$ and $y=\sqrt{u}$ $u=\sqrt{x}$ and $y=\dfrac{5}{u}$ or $u=\dfrac{25}{x}$ and $y=\sqrt{u}$ $\dfrac{dy}{dx}=-6(5-2x)^2$ $\dfrac{dy}{dx}=\dfrac{3}{2\sqrt{3x-1}}$ $\dfrac{dy}{dx}=\dfrac{-5}{2x^{3/2}}$ $y=9x^2-6x+1$ $y=e^{5x}$ $y=\ln{3x}$ $u=3x-1$ and $y=u^2$ $u=5x$ and $y=e^u$ or $u=e^x$ and $y=u^5$ $u=3x$ and $y=\ln{u}$ or $u=\ln{x}$ and $y=u+\ln{3}$ $\dfrac{dy}{dx}=6(3x-1)$ $\dfrac{dy}{dx}=5e^{5x}$ $\dfrac{dy}{dx}=\dfrac{1}{x}$ $y=\dfrac{1}{x^2+4x+4}$ $y=e^{x+4}$ $y=\ln{x^2}$ $u=x+2$ and $y=u^{-2}$ or $u=(x+2)^2$ and $y=\dfrac{1}{u}$ $u=x+4$ and $y=e^x$ or $u=e^x$ and $y=e^{4}u$ $u=x^2$ and $y=\ln{u}$ or $u=\ln{x}$ and $y=2u$ if $x>0$ $\dfrac{dy}{dx}=-2(x+2)^{-3}$ $\dfrac{dy}{dx}=e^{x+4}$ $\dfrac{dy}{dx}=\dfrac{2}{x}$ $y=\cot x$ $y=x$ $y=\ln{x^2}+\ln{8x}$ $u=\tan x$ and $y=\dfrac{1}{u}$ $u=x$ and $y=u$ or $u=e^x$ and $y=\ln{u}$ $u=8x^3$ and $y=\ln{u}$ or $u=2x$ and $y=3\ln{u}$ or $u=\ln{x}$ and $y=3u+3\ln{2}$ $\dfrac{dy}{dx}=-\cosec^2 x$ $\dfrac{dy}{dx}=1$ $\dfrac{dy}{dx}=\dfrac{3}{x}$ $y=\sin x^2$ $y=\tan x (\sec^2 x-1)$ $y=\tfrac{1}{2}(1-\cos 2x)$ $u=x^2$ and $y=\sin u$ $u=\tan x$ and $y=u^3$ $u=\cos 2x$ and $y=\tfrac{1}{2}-\tfrac{1}{2}u$ or $u=2x$ and $y=\tfrac{1}{2}-\tfrac{1}{2}\cos u$ or $u=\sin x$ and $y=u^2$ $\dfrac{dy}{dx}=2x\cos x^2$ $\dfrac{dy}{dx}=3\tan^2 x \sec^2 x$ $\dfrac{dy}{dx}=2\sin x \cos x=\sin 2x$