Chain rule for differentiation

Differentiation: Calculating derivatives and tangent lines

Chain rule for differentiation

With the chain rule we can determine the derivative of a composite function using the derivatives of the function from which it is composed.

Chain rule

For two functions $\blue{f(x)}$ and $\green{g(x)}$ holds $(\blue f \circ \green{g})'(x)=\blue f'(\green{g(x)}) \cdot \green{g'(x)}$

Example
$\frac{\dd}{\dd x} \blue ( \green{x^2-5x}\blue{)^4}=4 \cdot (\green{x^2-5x})^3 \cdot (2x-5)$

The chain rule states that the derivative of a composite function $\blue f(\green{g(x)})$ is equal to the derivative of the outer link $\blue f$ with the inner link $\green g$ inserted multiplied with the derivative of the inner link $\green g$ .
We call this a chain, because we can also consider a composite function as a chain function consisting of two links.

When applying the chain rule, it is important to first determine of which two functions $\blue{f(x)}$ and $\green{g(x)}$ the function is composed. In some cases there are multiple options possible. It is then important to choose these functions in such a way that both the derivative of $\blue{f(x)}$ and the derivative of $\green{g(x)}$ can be determined.

In the example the derivative of $(\blue f \circ \green g)(x)=\blue ( \green{x^2-5x}\blue{)^4}$ is determined. To do this, we first need to determine what the functions $\blue{f(x)}$ and $\green{g(x)}$ are.
In this case these functions are $\blue{f(x)}=\blue{x^4}$ and $\green{g(x)}=\green{x^2-5x}$ .
The derivatives of these functions are $f'(x)=4x^3$ and $g'(x)=2x-5$ respectively.
We can now find the derivative of $(\blue f \circ \green g)(x)=\blue ( \green{x^2-5x}\blue{)^4}$ using these derivatives and the chain rule, as is done in the example.

A different notation for the chain rule is
$\frac{\dd }{\dd x}\blue f(\green{g(x)})=\left.\frac{\dd }{\dd x}\blue{f(x)}\right|_{\green{g(x)}}\cdot \frac{\dd }{\dd x}\green{g(x)}$ Here $\left.\frac{\dd }{\dd x}f(x)\right|_{a}$ is used to represent $f'(a)$ , the derivative of $f(x)$ at the point $a$ .

Example
The function $\blue f(\green{g(x)})= \left(\frac{3}{x} +1\right)^2$ is composed of $\blue{f(x)}=\blue{x^2}$ and $\green{g(x)}=\green{\frac{3}{x}+1}$ . Then $\blue f(\green{g(x)})= \left(\frac{3}{x}+1\right)^2$ and thus: $\begin{array}{rcl}\dfrac{\dd}{\dd x}\left(\dfrac{3}{x}+1\right)^2&=&\left.\dfrac{\dd }{\dd x}\blue{x^2}\right|_{\green{\frac{3}{x}+1}}\cdot \dfrac{\dd }{\dd x}\left( \green{\dfrac{3}{x}+1} \right)\\&=& 2 \cdot \left( \green{\dfrac{3}{x} +1} \right) \cdot -\dfrac{3}{x^2} \\ &=& -\dfrac{18}{x^3} - \dfrac{6}{x^2}\end{array}$

The chain rule uses the derivative of $\green g$ in $x$ and the derivative of $\blue f$ in $\green{g(x)}$ . Both of these derivatives must exist to make the rule valid.

Let $\blue{f}$ and $\green{g}$ be two functions, where $\green g$ is differentiable and $\blue f$ is differentiable on the range of $g$ . Let $x$ be a point in the domain of $\green g$ .

To prove the chain rule, we will first define a new function $a(z)=\left\{\begin{array}{ll}\dfrac{\blue{f}(\green{g(x)}+z)-\blue f(\green{g(x)})}{z}-\blue f'(\green{g(x)}) & \text{if } z \ne 0 \\ 0 & \text{if } z=0\end{array}\right.$
Because of the definition of the derivative the following applies $\lim_{z \to 0}a(z)=\blue f'(\green{g(x)})-\blue f'(\green{g(x)})=0$
This means that $a(z)$ is continuous in $z=0$ . This will be used later on in this proof.

The definition of $a(z)$ for $z\ne 0$ can be rewritten by adding $\blue f'(\green{g(x)})$ to both sides and multiplying both sides with $z$ . This gives:
$\blue{f}(\green{g(x)}+z)-\blue f(\green{g(x)})=\left(\blue f'(\green{g(x)})+a(z)\right) \cdot z$
We will now substitute $z=\green{g(x+h)}-\green{g(x)}$ . This gives:
$\blue{f}(\green{g(x)}-(\green{g(x+h)}-\green{g(x)}))-\blue f(\green{g(x)})=\left(\blue f'(\green{g(x)}+a(\green{g(x+h)}-\green{g(x)})\right) \cdot \left(\green{g(x+h)}-\green{g(x)}\right)$
This can be simplified to
$\blue{f}(\green{g(x+h)})-\blue f(\green{g(x)})=\left(\blue f'(\green{g(x)}+a(\green{g(x+h)}-\green{g(x)})\right) \cdot \left(\green{g(x+h)}-\green{g(x)}\right)$
In the above expression we want to let $h$ go to $0$ . Therefore we will now determine $\lim_{h \to 0} a(\green{g(x+h)}-\green{g(x)})$ .
Because differentiable functions are always continuous, it holds that $\lim_{h \to 0}(\green{g(x+h)}-\green{g(x)})=0$ .
Earlier we showed that $a(z)$ is continuous in $z=0$ , therefore we can now use the calculation rule for composition of limits. This gives:
$\lim_{h \to 0}a(\green{g(x+h)}-\green{g(x)})=\lim_{z \to 0} a(z)=0$
Now we can determine the derivative of $\blue f(\green{g(x)})$ .
$\begin{array}{rcl}\dfrac{\dd }{\dd x}(\blue f(\green{g(x)}))&=& \displaystyle\lim_{h \to 0} \dfrac{\blue f(\green{g(x+h)})-\blue f(\green{g(x)})}{h} \\&&\quad \blue{\text{definition of derivative}}\\ &=& \displaystyle \lim_{h \to 0} \dfrac{\green{g(x+h)}-\green{g(x)}}{h} \cdot \left(\blue f'(\green{g(x)})+a(\green{g(x+h)}-\green{g(x)})\right) \\&&\quad \blue{\text{earlier derivation used}}\\ &=& \displaystyle \lim_{h \to 0} \dfrac{\green{g(x+h)}-\green{g(x)}}{h} \cdot \lim_{h\to 0} \left(\blue f'(\green{g(x)})+a(\green{g(x+h)}-\green{g(x)})\right) \\&&\quad \blue{\text{limits split as both exist}}\\ &=& \green{g'(x)} \cdot \left(\blue f'(\green{g(x)}+0\right) \\&&\quad \blue{\text{limits determined}}\\ &=& \green{g'(x)} \cdot \blue f'(\green{g(x)}) \\&&\quad \blue{\text{calculated}}\end{array}$
With this the chain rule is proven.

Determine the derivative of the function $\left(5\cdot x^2-3\right)^{{{1}\over{3}}}$ .

$\frac{\dd}{\dd x}\left (\left(5\cdot x^2-3\right)^{{{1}\over{3}}}\right)=$ ${{10\cdot x}\over{3\cdot \left(5\cdot x^2-3\right)^{{{2}\over{3}}} }}$

We can calculate $\dfrac{\dd}{\dd x}\left(\left(5\cdot x^2-3\right)^{{{1}\over{3}}}\right)$ by using the chain rule. Write $(f\circ g)(x)=\left(5\cdot x^2-3\right)^{{{1}\over{3}}}$ with $f(x)=x^{{{1}\over{3}}}$ and $g(x)=5\cdot x^2-3$ . Now e can apply the chain rule which states: $(f\circ g)'(x)=f'(g(x)) \cdot g'(x)$ .
$\begin{array}{rcl}\displaystyle \dfrac{\dd}{\dd x} \left(\left(5\cdot x^2-3\right)^{{{1}\over{3}}}\right) &=& \displaystyle {{1}\over{3\cdot g(x)^{{{2}\over{3}}}}} \cdot \frac{\dd }{\dd x}\left(g(x)\right) \\ &&\phantom{xxx}\blue{\text{chain rule applied with }f'(x)={{1}\over{3\cdot x^{{{2}\over{3}}}}}}\\ &=&\displaystyle \left({{1}\over{3\cdot \left(5\cdot x^2-3\right)^{{{2}\over{3}}}}}\right)\cdot \frac{\dd }{\dd x}\left(5\cdot x^2-3\right) \\ &&\phantom{xxx}\blue{g(x)=5\cdot x^2-3 \text{ substituted}}\\ &=&\displaystyle \left({{1}\over{3\cdot \left(5\cdot x^2-3\right)^{{{2}\over{3}}}}}\right)\cdot\left( 10\cdot x\right) \\ &&\phantom{xxx}\blue{\frac{\dd }{\dd x}\left(5\cdot x^2-3\right) \text{calculated}}\\ &=&\displaystyle {{10\cdot x}\over{3\cdot \left(5\cdot x^2-3\right)^{{{2}\over{3}}} }}\\ &&\phantom{xxx}\blue{\text{simplified}} \end{array}$

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