# The Chain Rule.

Chain Rule is one of the Techniques of Differentiation.

The Chain Rule states that:

If v(u) is a Function of “u” and u(x) is a function of “x” then  the Derivative of

function “v” with respect to “x” exists which  is equal to the product of derivative of function “v” with respect “u” and derivative of function “u” with respect to “x”.

Mathematically it can be written as:

If,  “v” is a function of “u” and “u” is a function of “x”,

$\frac{d}{dx}\left(v\right) = \frac{d}{du}\left(v\right) \times \frac{d}{dx}\left(u\right)$

Proof of The Chain Rule:

When, “u” is a function of “x” or $u=g(x)$ and “v” is a function of “u” or $v=f(u)$

Let, $\Delta x$ be a small increment in “x”. and $\Delta u$ Be the corresponding small increment in “u”.

So,
$u+ \Delta u =g(x+\Delta x)$

or,
$\Delta u=g(x+\Delta x)-g(x)$

So,
$\displaystyle \lim_{ \Delta x \to o} \Delta u = [g(x+ \Delta x)-g(x)]$

$=g(x)-g(x)=0$

So, $\Delta u\rightarrow o$ as $\Delta x\rightarrow o$. Then,

We know,
$\frac{\Delta v}{\Delta x}=\frac{\Delta v}{\Delta u} \times \frac{\Delta u}{\Delta x}$

So,
$\frac{d}{dx}\left(v\right) = \displaystyle \lim_{\Delta x\to o}\frac{\Delta v}{\Delta x}=\displaystyle \lim_{\Delta x\to o}\frac{\Delta y . \Delta u}{\Delta u . \Delta x}$

$=\left(\displaystyle\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta u}\right).\left(\displaystyle\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x}\right)$

Use of Chain Rule:

Find $\frac{d}{dx}\left(Y\right)$ if, $y=4u^2-3u+5$ and $u=2x^2-3$

Solution:

$\frac{d}{du}\left(Y\right)=8u-3$ and $\frac{d}{dx}\left(u\right)=4x$

So,
$\frac{d}{dx}\left(Y\right)=\frac{d}{du}\left(Y\right) . \frac{d}{dx}\left(u\right)$

$= (8u-3) 4x = [8(2x^2-3)-3] 4x = 64x^3-108x$

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