While finding the derivatives of a function we apply different rules like product rule, quotient rule or chain rule. These rules are applied according to the types of functions.

Usually the functions are mentioned in explicit form $y$ = $f\ (x)$ or in the implicit form $f\ (x,\ y)$ = $constant$.

The chain rule is applied for the functions which are expressed in explicit form. Initially we understand that the function is in composite form and then split it into different functions and then apply chain rule.
The chain rule can be applied for the exponential functions, logarithmic functions, inverse trigonometric functions etc.
In this section let us see the proof and the method of finding the derivatives using chain rule. We also have some practice problems and their Answers at the end of this unit.

## Definition

If $f\ (x)$ is a real valued function and if $h$ is a small increment given to $h$, then the derivative $f'\ (x)$ is given by the limit,

$f'(x)$ = $\lim_{h\to0}\frac{f(x+h)-f(h)}{h}$

## Chain Rule Proof

Chain Rule for Derivatives:

We can define the chain rule for a composite function as follows.

If $y$ = $(f\ o\ g)\ (x)$ is a real valued function,

Then the derivative of $(f\ o\ g)\ (x)$ is given by,

$\frac{dy}{dx}$
=
$\lim_{h\to0}\frac{(fog)(x+h)-(fog)(x)}{h}$.

The above function can also be written as $y$ = $f\ (u)$ and $u$ = $g\ (x)$

then the derivative $\frac{dy}{dx}$ =
$\frac{dy}{du}$ . $\frac{du}{dx}$
Chain Rule Proof:

Let $(f\ o\ g)\ (x)$ be a real valued function which is continuous.

Let h be a small increment given to $x$.

Then the according to the definition of derivatives,

Derivative of $(f\ o\ g)\ (x)$ is given by,

$\frac{dy}{dx}$ = $\lim_{h\to0}\frac{(fog)(x+h)-(fog)(x)}{h}$

Therefore, $\frac{dy}{dx}$ =
$\lim_{h\to0}\frac{(f[g(x+h)]-(f[g(x)]}{g(x+h)-g(x)}$.$\frac{g(x+h)-g(x)}{h}$ [multiplying and dividing by $g (x\ +\ h)\ -\ g\ (x)$ ]

$(f\ o\ g)'\ (x)$ = $f'\ (g\ (x))\ .\ g'\ (x)$
Multivariable Chain Rule:

If $y$ = $f\ (x)$ is such that we write the functions interms of $u,\ v$ and $w$,

Where $y$ = $f\ ( u ),\ u$ = $g\ (v)$ and $v$ = $h\ (x)$

Then $\frac{dy}{dx}$ =
$\frac{dy}{du}$ . $\frac{du}{dv}$ . $\frac{dv}{dx}$

## Chain Rule Examples

Example 1:

Differentiate $h(x)$ when $h(x)$ = $f(g(x))$ and it is given that $g(x)$ = $2\ -\ 3x$ and $f(x)$ = $2x^{2}$.

Solution:

Using the chain rule of differentiation we have,

$h'(x)$ = $f'(g(x)g'(x)$

Differentiating both functions we get, $g'(x)$ = $-3$ and $f'(x)$ = $4x$

$h'(x)$ = $4(g(x))$ $\times\ (-3)$ = $4(2\ -\ 3x)(-3)$ = $-12(2\ -\ 3x)$ = $36x$ - $24$.
Example 2:

If $g(x)$ = $e^{x}$ and $f(x)$ = $5x$ find the derivative of $f(g(x))$.

Solution:

Differentiating we get, g'(x) = $e^{x}$ and f'(x) = 5.

Using the chain rule for derivatives we have,

$f'(g(x))$ = $f'(g(x))g'(x)$ = $f'(e^{x})e^{x}$ = $5e^{x}\ e^{x}$ = $5e^{2x}$

### Solved Examples

Question 1: Differentiate with respect to x.
f (x) = $\sqrt{[\left (x^{2} +a^{2}\right)^{n}]}$
Solution:

We have f (x) = $\sqrt{[ \left (x^{2} +a^{2}\right)^{n}]}$
Let y = f (u)
= $\sqrt{u}$
where u = vn
and v = $\left (x^{2} +a^{2}\right)$
Therefore,

$\frac{dy}{du}$ = $\frac{d}{du}$ [ $\sqrt{u}$ ]

= $\frac{1}{2}$ . u(1/2) - 1

= $\frac{1}{2}$ . u-1/2

= $\frac{1}{2}$ . $\frac{1}{\sqrt{u}}$

$\frac{du}{dv}$ = $\frac{d}{dv}$ [ vn ]

= n vn-1

$\frac{dv}{dx}$ = $\frac{d}{dx}$ . [ $\left (x^{2} +a^{2}\right)$ ]

= 2 x

Multiplying the above derivatives, we get,
$\frac{dy}{dx}$ = $\frac{dy}{du}$ . $\frac{du}{dv}$ . $\frac{dv}{dx}$

=
$\frac{1}{2}$ . $\frac{1}{\sqrt{u}}$ . n vn-1 . 2 x

= n vn-1 x . $\frac{1}{\sqrt{v^{n}}}$

Question 2: Suppose that F(x) = f (g (x)) and g (2) = 8, g'(2) = 10, f'(2) = 4 and f'(8) = 12, find F'(2).
Solution:

We have g (2) = 8, g'(2) = 10, f'(2) = 4 and f'(8) = 12
According to chain rule, if F (X) = (f o g) (x)
then F'(x) = f'(g (x)) . g'(x)
substituting x = 2, we get,
F'(2) = f'(g (2)) . g'(2) = f'(8) . 10 = 12 . 10 = 120
Therefore, F'(2) = 120

Question 3: Differentiate esec 3x with respect to x.
Solution:

Let y = f (x) = esec 3x
Applying chain rule directly

$\frac{dy}{dx}$ = $\frac{d}{dx}$ esec 3x

= esec 3x . $\frac{d}{dx}$ [ sec 3x ]

= esec 3x . sec 3x . tan 3x . $\frac{d}{dx}$ [ 3 x ]

f'(x) = 3 . sec 3x . tan 3x . esec 3x

## Chain Rule Practice Problems

### Practice Problems

Question 1: sin (3x + 5)
Question 2: cos (ln x)
Question 3: e 3x . cos 2x
Question 4: earc sin 2x
Question 5: cos (ln x) 2
Question 6: log (csc x - cot x)
Question 7: x sin 2x + 5x + kk + ( tan2 x )3
Question 8: log (cos x2)
Question 9: If y = log { $\sqrt{(x-1)}$ - $\sqrt{(x+1)}$ }, show that $\frac{dy}{dx}$ = $\frac{-1}{2\sqrt{(x^{2}-1)}}$
Question 10: If y = $\sqrt{x^{2}-a^{2}}$, prove that y . $\frac{dy}{dx}$ + x = 0