Derivatives of Composite Functions: Chain Rule, Generalisation, Examples

Sometimes complex looking functions can be greatly simplified by expressing them as a composition of two or more different functions. It is then not possible to differentiate them directly as we do with simple functions. In this topic, we shall discuss the differentiation of such composite functions using the Chain Rule. We’ll discuss the rule with proof for the composition of two functions only, but it can be extended to functions involving multiple compositions as well. So, let’s begin!

The Chain Rule

Let f(x) and g(x) be two differentiable functions with a common domain. Then the derivative of a function formed by a composition of these two functions i.e. f(g(x)) is given by –

$$ {\frac{d}{dx}f(g(x)) = f'(g(x)).g'(x)} $$

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It basically asks you to first differentiate the outer function i.e. f(x) normally, without touching the inner function. Then multiply the result with the derivative of the inner function i.e. g(x) here. Let us prove this now.

Chain Rule

Defining the Chain Rule

To begin with, let us introduce a variable u = g(x) to simplify the looks of our steps.
Note that now in terms of this new variable, we need to prove:
$$ {\frac{d}{dx}f(u) = f'(u)\frac{du}{dx}} $$

Since u = g(x), it is obviously a function of x; and since we have assumed g(x) to be differentiable, u will also be differentiable. Then using the definition of the derivative, we can write u'(x) as:
$$ {u'(x) = Lim_{x\rightarrow{0}}\frac{u(x + h) – u(x)}{h}} $$

Let us now define a function v(h) as:
$${v(h) = \frac{u(x + h) – u(x)}{h} – u'(x) \text{ if h}\neq 0} $$
$$ {\text{and }v(h) = 0 \text{ if h = 0}} $$

Do understand the significance of us defining the function at x = 0. Check for the limit h → 0 in the function v(h). We’ll get –

$${Lim_{h\rightarrow{0}} v(h) = Lim_{h\rightarrow{0}}(\frac{u(x + h) – u(x)}{h}) – Lim_{h\rightarrow{0}}u'(x)} $$

Thus, $${ Lim_{h\rightarrow{0}} v(h) = u'(x) – u'(x) = 0}$$ following from the definition of u'(x). This makes the function v(h) continuous at every point in its domain including h = 0 since Lim_{h → 0}v(h) = 0 = v(h = 0). Now, if we consider that h ≠ 0, we can manipulate the definition of v(h) for h ≠ 0 to get:

${ u(x + h) = u(x) + h.(v(h) + u'(x))}$

Note that this equation is valid at h = 0 as well! Thus, this function is valid for all h. We can do the same job for the function f(x) now:

$${w(k) = \frac{f(z + k) – f(z)}{k} – f'(z) \text{ if k}\neq 0} $$
$$ {\text{and }w(k) = 0 \text{ if k = 0}} $$

It then follows similarly:

${ f(z + k) = f(z) + k.(w(k) + f'(z))}$

This much analysis is sufficient for us to now get started on the proof.

Proof:

Using the definition of the derivative, we have:

$$\frac{d}{dx}[f\:[u(x)]]\:= \:Lim _{h\rightarrow{0}}\:\frac{f[u(x+h)] \: – \: f[u(x)]}{h}$$

Using the equation we obtained for u(x + h), we can get the numerator in the right-hand side as:

$f[u(x+h)]-f[u(x)]\:=\:f[u(x)+h(v(h)+u'(x))]-f[u(x)] $

If we use $ z=u(x)$, and $k=h(v(h)+u'(x))$ in our equation for f(z + h) above, we can further get the numerator as;

$f[u(x+h)]-f[u(x)]\:=\:f[u(x)+h(v(h)+u'(x))]-f[u(x)] $

$ =\:f[u(x)]+h(v(h)+u'(x))\:(w(k)+f'[u(x)]) – f[u(x)]$

$=h(v(h)+u'(x))\:(w(k)+f'[u(x)]$)

Putting it back in the limit on the right-hand side, we now actually have –

$$\frac{d}{dx}[f[u(x)]]\:=\:Lim_{h\rightarrow{0}}\frac{h(v(h)+u'(x))(w(k)+f'[u(x)])}{h}$$

The h cancels out from the numerator and the denominator giving us –

$=Lim_{h\rightarrow{0}}(v(h)+u'(x))(w(k)+f'[u(x)])$

Now, if you recall; we had defined k as $k=h(v(h)+u'(x))$. Taking its limit as h → 0 :

$Lim_{h\rightarrow{0}}k=Lim_{h\rightarrow{0}}h(v(h)+u'(x))=0$

Using the definition of the function w(k) and the fact that w(k) is continuous at k = 0, we then have –

$Lim_{h\rightarrow{0}}w(k) = w(Lim_{h\rightarrow{0}}k)=w(0)=0$

Also, known from above – $Lim_{x\rightarrow{0}}v(h) = 0$. Using these facts, we finally have the limit on the right-hand side as –

$$\frac{d}{dx}[f[u(x)]] = Lim_{h\rightarrow{0}}(v(h)+u'(x))(w(k)+f'[u(x)])$$

$=u'(x)f'[u(x)]$

$${=f'[u(x)]\frac{du}{dx}}$$

This was what we had to prove! Thus, the Chain Rule is obtained.

Generalisation of the Chain Rule

Here, we will give you the formula for finding the derivatives of the functions that involve the composition of multiple simple functions. The form of this general Chain Rule is very simple to understand if you understood the Chain Rule for the composition of two simple functions. Let us now look at it for a function which is a composition of more than two functions f(g(h(…p(x)))) –

$$ {\frac{d}{dx}f(g(h(…p(x)))) = [f'(g(h(…p(x))))].[g'(h(…p(x)))].[h'(…p(x))]…[p'(x)]} $$

Now take a look at the solved examples below to better identify composite functions and clear your concepts on applying the chain rule!

Solved examples for You

Question 1: Find the derivative of ${sin(\frac{1}{x})}$.

Answer : This is a composition of two functions: f(x) = sin x and g(x) = ${\frac{1}{x}}$. Clearly the form of composition here is f(g(x)). By the Chain Rule, we then have –

$ {\frac{d}{dx}sin(\frac{1}{x}) = cos(\frac{1}{x}).\frac{d}{dx}(\frac{1}{x})} $

$ { = cos(\frac{1}{x}).(-\frac{1}{x^2})} $

$ { = (-\frac{cos(\frac{1}{x})}{x^2})} $

which is the required result.

Question 2: Find the derivative of ${\frac{1}{sin(x^2)}}$.

Answer : This is a composition of three functions given below:
f(x) = ${\frac{1}{x}}$
g(x) = sin x
h(x) = x²

The form of the function given to us is: f(g(h(x))). By the General Chain Rule, we then have its derivative as –

${ \frac{d}{dx}(\frac{1}{sin(x^2)}) = [-\frac{1}{(sin(x^2))^2}].[cos(x^2)].[2x] }$

${ = -\frac{2x.cos(x^2)}{(sin(x^2))^2} }$

which is the required result.

Question 3: Prove the Quotient Rule of Differentiation as a consequence of the Chain Rule on the Product Rule of Differentiation.

Answer : Let us have a function $\frac{f(x)}{g(x}$. Its derivative can be written by the product rule as –
$${\frac{d}{dx}\frac{f(x)}{g(x)} = f'(x).\frac{1}{g(x)} + \frac{d}{dx}(\frac{1}{g(x)}).f(x) } $$

Now look at the derivative $ \frac{d}{dx}(\frac{1}{g(x)})$. It can be considered as a derivative of the composition of the following functions – g(x) and p(x) = $\frac{1}{x}$. Thus its derivative can be written by the Chain Rule as –
$${ \frac{d}{dx}(\frac{1}{g(x)}) = (-\frac{1}{(g(x))^2}).g'(x) } $$

Using this derivative back to get the Quotient Rule :
$${\frac{d}{dx}\frac{f(x)}{g(x)} = f'(x).\frac{1}{g(x)} + \frac{d}{dx}(\frac{1}{g(x)}).f(x) } $$

$ = f'(x).\frac{1}{g(x)} – (\frac{1}{(g(x))^2}).g'(x).f(x)$

$ = \frac{f'(x).g(x) – g'(x).f(x)}{(g(x))^2}$

which proves the Quotient Rule of Differentiation.

This concludes our discussion on the Derivative of Composite Functions.

Question 4: Explain the chain rule formula?

Answer: The chain rule explains that the derivative of f(g(x)) is f'(g(x))⋅g'(x). In other words, the chain rule helps in differentiating *composite functions*. For example, sin(x²) is a composite function due to the fact that its construction can take place as f(g(x)) for f(x)=sin(x) and g(x)=x².

Question 5: Why is chain rule workable?

Answer: There is a reason for the workability of simple form of the chain rule for linear functions. The reason is that the derivatives are constants. Furthermore, the derivatives are independent of the inputs value to the functions. While making use of the chain rule, one must pay attention to the evaluation of the derivative of f at g′(x). Also, one must make use of the valid chain rule h′(x)=f′(g(x))g′(x).

Question 6: What is meant by dx?

Answer: Literally speaking, dx refers to an infinitely small length belonging to x.

Question 7: What is the origin of chain rule?

Answer: The chain rule has been in existence since the discovery of calculus by Isaac Newton and Leibniz at the end of the 17th century. The rule facilitates calculations that deal with derivatives of complex expressions.