21.6 - The Chain Rule

In the previous section, we completed a discussion on differentiability. In this section, we will see derivatives of composite functions.

First we will see an example. It can be written in 3 steps:
1. We want the derivative of f, where f(x) = (3x+4)³.
2. Let us expand (3x+4)³. We get:
f(x) = 27x³ + 108 x² + 144 x + 64
3. This is a polynomial function. So we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d}{dx} {f(x)}}    & {~=~}    &{\frac{d}{dx} {\left[(3x+4)^3 \right]}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{d}{dx} \left(27 x^3 + 108 x^2 + 144x + 64 \right)}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{27(3)x^2 + 108 (2) x + 144 + 0}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{81x^2 + 216 x + 144 }    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{9 \left(9 x^2 + 24 x + 16 \right)}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{9 (3x + 4)^2}    \\ \end{array}$

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Now we will see an easier method. The details about this new method can be written in 9 steps:

1. Consider the composite function f(x) = (v∘u)(x). It can also be written as f(x) = v(u(x)).
• We see that, the output of u is being used as the input for v.
2. The function in our example is: f(x) = (3x+4)³.
We can write it as the composite of two functions: f(x) = v(u(x))
    ♦ Where u(x) = 3x+4 and v(u(x)) = [u(x)]³.
• Then f(x) = v(u(x)) = v(3x+4) =  (3x+4)³.
3. Put u(x) = 3x+4 = t.
Then we get: f(x) = v(t) = t³ = (3x+4)³.
4. Recall the derivative that we obtained above, by considering it as a polynomial function.
• We obtained 9(3x+4)².
• So now we can write:
$\frac{d}{dx} {f(x)} \,=\, 9 (3x+4)^2 \,=\, \left[3 (3x+4)^2 \right].3\,=\, \left[3 t^2 \right].3$
5. Let us write the general form for 3t² and 3.
• We wrote: v(t) = t³.
   ♦ So 3t² is $\frac{dv}{dt}$
• Similarly we wrote: t = 3x+4.
   ♦ So 3 is $\frac{dt}{dx}$
6. Now the result in (4) becomes:
$\frac{d}{dx} {f(x)} \,=\, 9 (3x+4)^2 \,=\, \left[3 (3x+4)^2 \right].3\,=\, \frac{dv}{dt}.\frac{dt}{dx}$
• Taking the first and last terms, we get:
$\frac{df}{dx} \,=\, \frac{dv}{dt}.\frac{dt}{dx}$
7. The result obtained in the above step (6) is known as the chain rule.
• Let us write the 4 steps for applying this rule:
(i) We are given the composite function: f(x) = (v∘u)(x)
• This can be written as f(x) = v(u(x))
• So output of u is the input of v. That means, u can be considered as the inner function.
(ii) We put inner function u(x) = t
From this, we can find $\frac{dt}{dx}$
(iii) Write the outer function v in terms of t.
From this, we can find $\frac{dv}{dt}$
(iv) Finally we can calculate $\frac{df}{dx}$ as follows:
$\frac{df}{dx} \,=\, \frac{dv}{dt}.\frac{dt}{dx}$
8. Chain rule can be extended to composite of 3 functions.
Let us write the 5 steps:
(i) We are given the composite function: f = w∘u∘v
(ii) We put the inner most function v = t
From this, we can find $\frac{dt}{dx}$
(iii) The next inner function is u.
• Write u in terms of t. This gives u(t)
• Put u(t) = s. From this, we can find $\frac{ds}{dt}$
(iv) The outer function is w.
• Write w in terms of s. This gives w(s)
• From this, we can find $\frac{dw}{ds}$
(v) Finally we can calculate $\frac{df}{dx}$ as follows:
$\frac{df}{dx} \,=\, \frac{dw}{ds}.\frac{ds}{dt}.\frac{dt}{dx}$
9. In this way, we can extend the chain rule to composite of 4 or more functions.

Solved example 21.21
Find the derivative of the function given by f(x) = sin (x²)
Solution:
1. f(x) = sin (x²) = (v∘u)(x) = v(u(x))
Where u(x) = x² and v(u(x)) = sin (u(x)).
2. Put u(x) = x² = t.
So $\frac{dt}{dx} = 2x$
3. v(u(x)) = sin (u(x))
• So v(t) = sin t
⇒ $\frac{dv}{dt} = \cos t = \cos (x^2)$
4. $\frac{df}{dx} = \frac{dv}{dt} . \frac{dt}{dx} = \cos (x^2) . 2x = 2x \cos (x^2)$

Solved example 21.22
Find the derivative of the function given by f(x) = tan (2x+3)
Solution:
1. f(x) = tan (2x+3) = (v∘u)(x) = v(u(x))
Where u(x) = 2x+3 and v(u(x)) = tan (u(x)).
2. Put u(x) = 2x+3 = t.
So $\frac{dt}{dx} = 2$
3. v(u(x)) = tan (u(x))
So v(t) = tan t
⇒ $\frac{dv}{dt} = \sec^2 t = \sec^2 (2x+3)$
4. $\frac{df}{dx} = \frac{dv}{dt} . \frac{dt}{dx} = \sec^2 (2x+3) . 2 = 2 \sec^2 (2x+3)$

Solved example 21.23
Find the derivative of the function given by f(x) = sin (cos (x²))
Solution:
1. f(x) = sin (cos (x²)) = (w∘v∘u)(x) = (w∘v)(u(x))
Where u(x) = x²,  v(u(x)) = cos (u(x)) and w(v(u(x))) = sin (v(u(x))).
2. Put u(x) = x² = t.
So $\frac{dt}{dx} = 2x$
3. Put v(u(x)) = cos (u(x)) = cos t = s.
So $\frac{ds}{dt} = - \sin t = -\sin x^2$
4. w(v(u(x))) = sin (v(u(x)))
so w(s) = sin (s)
⇒ $\frac{dw}{ds} = \cos s = \cos (\cos t) = \cos (\cos x^2)$
5. $\frac{df}{dx} \,=\, \frac{dw}{ds}.\frac{ds}{dt}.\frac{dt}{dx}$
= cos (cos x²). −sin x². 2x
= −2x sin x². cos (cos x²)

Solved example 21.24
Find the derivative of the function given by
$f(x) = \sqrt{5x - 8}$
Solution:
1. $f(x) = \sqrt{5x - 8}$ = (v∘u)(x) = v(u(x))
Where u(x) = 5x−8 and v(u(x)) = $\sqrt{u(x)}$.
2. Put u(x) = 5x−8 = t.
So $\frac{dt}{dx} = 5$
3. v(u(x)) = $\sqrt{u(x)}$
So v(t) = $\sqrt{t}$
⇒ $\frac{dv}{dt} = \frac{1}{2} (t)^{- \frac{1}{2}} = \frac{1}{2 \sqrt{5x - 8}}$
4. $\frac{df}{dx} = \frac{dv}{dt} . \frac{dt}{dx} = \frac{1}{2 \sqrt{5x - 8}} . 5 = \frac{5}{2 \sqrt{5x - 8}}$

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We have seen the basics about chain rule. In the next section, we will see an alternate method for applying this rule.

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