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 + $sin(x)^{2}$ |
$\frac{\text{d}}{\text{d}x} [(f O g)(x)] = \frac{\text{d}}{\text{d}g(x)}[(f O g)(x)].\frac{\text{d}}{\text{d}x}((g(x))$
$\frac{\text{d}y}{\text{d}x}= \frac{\text{d}y}{\text{d}u}.\frac{\text{d}u}{\text{d}x}$
$ \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 $\lim_{\triangle x \rightarrow 0}(x + \triangle x) = g(x))$
$\frac{\text{d}}{\text{d}x} [(f O g)(x)] = \frac{\text{d}}{\text{d}g(x)}[(f O g)(x)].\frac{\text{d}}{\text{d}x}((g(x))$
$\frac{\text{d}y}{\text{d}x}= \frac{\text{d}y}{\text{d}u}.\frac{\text{d}u}{\text{d}x}$
$\frac{\text{d}y}{\text{d}u}= cos(u).2x$
= 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) $