Chain Rule

Example 1: Using the Chain Rule

Let $f(textcolor{blue}{x}) = sqrt{2textcolor{blue}{x} – dfrac{1}{textcolor{blue}{x}^2}}$. Find the derivative of $f(textcolor{blue}{x})$ with respect to $textcolor{blue}{x}$.

[4 marks]

Take $g(textcolor{red}{u}) = sqrt{textcolor{red}{u}}$ and $textcolor{red}{u} = h(textcolor{blue}{x}) = 2textcolor{blue}{x} – dfrac{1}{textcolor{blue}{x}^2}$, so $f(textcolor{blue}{x}) = g(h(textcolor{blue}{x}))$.

Then,

$dfrac{dg(textcolor{red}{u})}{dtextcolor{red}{u}} = dfrac{1}{2sqrt{textcolor{red}{u}}} = dfrac{1}{2sqrt{2textcolor{blue}{x} – dfrac{1}{textcolor{blue}{x}^2}}}$ and $dfrac{dtextcolor{red}{u}}{dtextcolor{blue}{x}} =dfrac{dh(textcolor{blue}{x})}{dtextcolor{blue}{x}} = 2 + dfrac{2}{textcolor{blue}{x}^3} = dfrac{2textcolor{blue}{x}^3 + 2}{textcolor{blue}{x}^3}$.

Multiplying the two gives

$dfrac{df(textcolor{blue}{x})}{dtextcolor{blue}{x}} = left( dfrac{2textcolor{blue}{x}^3 + 2}{textcolor{blue}{x}^3}right) left( dfrac{1}{2sqrt{2textcolor{blue}{x} – dfrac{1}{textcolor{blue}{x}^2}}}right) = dfrac{textcolor{blue}{x}^3 + 1}{textcolor{blue}{x}^3 sqrt{2textcolor{blue}{x} – dfrac{1}{textcolor{blue}{x}^2}}}$

Example 2: Connecting Rates of Change

The surface area of a cube of width $x text{ cm}$, length $2x text{ cm}$ and height $4x text{ cm}$ decreases at a constant rate of $0.07 text{ cm}^2 text{s}^{-1}$. Find $textcolor{purple}{dfrac{dx}{dt}}$ at the point where $textcolor{orange}{x = 5 text{ cm}}$.

[3 marks]

The surface area of the cuboid is:

$A = 2 times (x times 2x) + 2 times (x times 4x) + 2 times (2x times 4x)  = 28x^2$

So,

$dfrac{dA}{dx} = 56x$

$A$ decreases at a constant rate of $0.07 text{ cm}^2text{s}^{-1}$, so we can write this as

$dfrac{dA}{dt} = – 0.07$

Now, use the chain rule to find $textcolor{purple}{dfrac{dx}{dt}}$:

$begin{aligned} textcolor{purple}{dfrac{dx}{dt}} &= dfrac{dx}{dA} times dfrac{dA}{dt} [1.2em] &= dfrac{1}{left( dfrac{dA}{dx} right)} times dfrac{dA}{dt} [1.2em] &= dfrac{1}{56x} times (-0.07) [1.2em] &= – dfrac{0.00125}{x} end{aligned}$

When $textcolor{orange}{x = 5}$,

$textcolor{purple}{dfrac{dx}{dt}} = – dfrac{0.00125}{x} = – dfrac{0.00125}{textcolor{orange}{5}} = -0.00025 text{ cms}^{-1}$