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 \(y=u^2\) |

\(\dfrac{dy}{dx}=2x\cos x^2\) |
\(\dfrac{dy}{dx}=3\tan^2 x \sec^2 x\) |
\(\dfrac{dy}{dx}=2\sin x \cos x=\sin 2x\) |