Question

How will you prove the chain rule of derivative using first principle?​

Answer 1

$\large\underline{\sf{Solution-}}$

Definition of Chain Rule :

If f and g are differentiable functions, then f o g is also differentiable and is represented as

$\rm :\longmapsto\:\boxed{ \tt{ \: (fog)'(x) = f'(g(x)) × g'(x) \: }}$

Proof :-

Since, f and g are differentiable function, so

By definition of First Principal, we have

$\rm :\longmapsto\:f'(x) \: = \: \displaystyle\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

and

$\rm :\longmapsto\:g'(x) \: = \: \displaystyle\lim_{h \to 0} \frac{g(x + h) - g(x)}{h}$

Now, Consider

$\rm :\longmapsto\:(fog)'(x) \: = \: \displaystyle\lim_{h \to 0} \frac{(fog)(x + h) -(fo g)(x)}{h}$

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{f[g(x + h)] - f[g(x)]}{h}$

can be rewritten as

• after multiply and divide by g(x + h) - g(x),

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{f[g(x + h)] - f[g(x)]}{g(x + h) - g(x)} \times \frac{g(x + h) - g(x)}{h}$

$\rm \: = \:\displaystyle\lim_{h \to 0} \frac{f[g(x + h)] - f[g(x)]}{g(x + h) - g(x)} \times \displaystyle\lim_{h \to 0}\frac{g(x + h) - g(x)}{h}$

$\rm \: = \:f'[g(x) \times g'(x)$

Hence,

$\red{\rm \implies\:\:\boxed{ \bf{ \: (fog)'(x) = f'(g(x)) × g'(x) \: }}}$

More to know : -

$\green{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$