Chain Rule for Differentiation

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The chain rule for differentiation deals with the composite functions.
For example compare the functions below :
Without the chain rule
1) $x^{3}+1$
2) cos(x)
3) (2x+1)
4) x + sin(x)
With the chain rule
1) $\sqrt{x^{3}+1}$
2) cos(4x)
3) $(2x +1)^{4}$
4) x + $sin(x)^{2}$
Basically, the Chain Rule states that if 'y' changes dy/du times as fast as 'u' and 'u' changes du/dx times as fast as 'x' then 'y' changes (dy/du)(du/dx) times as fast as 'x'.

Theorem : (chain rule) If f(x) or u and g(x) or v are differentiable functions, then (f O g) is also differentiable and
(f O g)' = f'(g(x)). g'(x)


$\frac{\text{d}}{\text{d}x} [(f O g)(x)] = \frac{\text{d}}{\text{d}g(x)}[(f O g)(x)].\frac{\text{d}}{\text{d}x}((g(x))$

$\frac{\text{d}y}{\text{d}x}= \frac{\text{d}y}{\text{d}u}.\frac{\text{d}u}{\text{d}x}$

Proof :
Since f(x) and g(x) are the differentiable functions. Therefore, according to the definition of derivative

$ \frac{\text{d}}{\text{d}x} [(f(x)] =\lim_{\triangle x \rightarrow 0}\frac{f(x+\triangle x) -f(x)}{\triangle x}$

$ \frac{\text{d}}{\text{d}x} [g(x)] =\lim_{\triangle x \rightarrow 0}\frac{g(x+\triangle x) -f(x)}{\triangle x}$

Now, $\frac{\text{d}}{\text{d}x} [(f O g)(x)]=\lim_{\triangle x \rightarrow 0}\frac{fOg(x+\triangle x) -fOg(x)}{\triangle x}$

$=\lim_{\triangle x \rightarrow 0}\frac{f[g(x+\triangle x)] -f[g(x)]}{\triangle x}$

$=\lim_{\triangle x \rightarrow 0}\frac{f[g(x+\triangle x)] -f[g(x)].[g(x + \triangle x) - g(x)]}{\triangle x.[g(x + \triangle x) - g(x)]}$

$=\lim_{\triangle x \rightarrow 0}\frac{f[g(x+\triangle x)] -f[g(x)]}{[g(x + \triangle x) - g(x)]}.\lim_{\triangle x \rightarrow 0}\frac{g(x + \triangle x) - g(x)}{\triangle x}$

$=\lim_{g(x+h) \rightarrow g(x)}\frac{f[g(x+\triangle x)] -f[g(x)]}{[g(x + \triangle x) - g(x)]}.\lim_{\triangle x \rightarrow 0}\frac{g(x + \triangle x) - g(x)}{\triangle x}$

(since g(x) is differentiable, g(x) is continuous and hence $\lim_{\triangle x \rightarrow 0}(x + \triangle x) = g(x))$

$\frac{\text{d}}{\text{d}x} [(f O g)(x)] = \frac{\text{d}}{\text{d}g(x)}[(f O g)(x)].\frac{\text{d}}{\text{d}x}((g(x))$

Examples on Chain Rule for Differentiation

Example 1: Differentiate with respect to x
y= sin($x^{2} +1 ) $
Solution : y= sin($x^{2} +1 ) $
Let u = $x^{2} +1 $
y= sin(u) and u =$x^{2} +1 $
$\frac{\text{d}y}{\text{d}u}= cos(u)$ and $\frac{\text{d}u}{\text{d}x}= 2x $

Now by chain rule for differentiation,

$\frac{\text{d}y}{\text{d}x}= \frac{\text{d}y}{\text{d}u}.\frac{\text{d}u}{\text{d}x}$

$\frac{\text{d}y}{\text{d}u}= cos(u).2x$

= 2x cos($x^{2}+1)$

$\frac{\text{d}}{\text{d}x}[sin(x^{2} +1)] = 2x cos(x^{2} +1) $

Example 2: Differentiate with respect to x
y= ($e^{sin(x)} ) $
Solution : y= ($e^{sin(x)} ) $
Let u = sin(x)
y= $e^{u}$ and u = sin(x)
$\frac{\text{d}y}{\text{d}u}=e^{u} $ and $\frac{\text{d}u}{\text{d}x}= cos(x) $

Now by chain rule for differentiation,

$\frac{\text{d}y}{\text{d}x}= \frac{\text{d}y}{\text{d}u}.\frac{\text{d}u}{\text{d}x}$

$\frac{\text{d}y}{\text{d}u}= e^{u}.cos(x)$

= $e^{sin(x)}.cos(x)$

$\frac{\text{d}}{\text{d}x}[(e^{sin(x)} )] = e^{sin(x)}.cos(x) $

12th grade math


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