#
Chain Rule for Differentiation

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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 + $sin(x)^{2}$ |

**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)

OR

(f O g)' = f'(g(x)). g'(x)

OR

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

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

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

Proof :

Proof :

$ \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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