3.4 Chain Rule

3.1 Scalar Differentiation:
3.2 Product Rule:
3.3 Quotient Rule:
3.4 Chain Rule

4.1 Derivative of Trigonometric functions
4.2 Derivative of Inverse Trigonometric functions
4.3 Derivative of Exponential and Logarithmic Function
4.4 Derivative Of Hyperbolic function
4.5 Questions related to derivation of common functions

8.1 Differentiation of determinant matrix

10.1 Questions related to Differentiation using substitution

$$\left( {fog(x)} \right)' = g'(x)f'(g(x))$$$$\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}$$ Proof :

$$ {\left( {fog\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to o} \frac{{fog\left( {x + h} \right) - fog\left( x \right)}}{h} $$$$ {\left( {fog\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to o} \frac{{f\left( {g\left( {x + h} \right)} \right) - f\left( {g\left( x \right)} \right)}}{h} $$

Since $g$ is differentiable at $x$,

$$ v = \frac{{g\left( {x + h} \right) - g\left( x \right)}}{h} - g'\left( x \right) $$$$ g\left( {x + h} \right) = g\left( x \right) + \left( {v + g'\left( x \right)} \right)h $$

with $$\mathop {\lim }\limits_{h \to 0} v = 0$$ Similarly,

$$ f\left( {y + h} \right) = f\left( y \right) + \left( {w + f'\left( y \right)} \right)k $$

with $$\mathop {\lim }\limits_{k \to 0} w = 0$$

In particular $y = g\left( x \right)$ and $k = \left( {v + g'\left( x \right)} \right)h$

$$ f\left( {g\left( x \right) + \left( {v + g'\left( x \right)} \right)h} \right) = f\left( {g\left( x \right)} \right) + \left( {w + f'\left( {g\left( x \right)} \right)} \right)\left( {v + g'\left( x \right)} \right)h $$

Hence,

$$ f\left( {g\left( {x + h} \right)} \right) - f\left( {g\left( x \right)} \right) = f\left( {g\left( x \right) + \left( {v + g'\left( x \right)} \right)h} \right) - f\left( {g\left( x \right)} \right) $$$$ f\left( {g\left( {x + h} \right)} \right) - f\left( {g\left( x \right)} \right) = f\left( {g\left( x \right)} \right) + \left( {w + f'\left( {g\left( x \right)} \right)} \right)\left( {v + g'\left( x \right)} \right)h - f\left( {g\left( x \right)} \right) $$$$ f\left( {g\left( {x + h} \right)} \right) - f\left( {g\left( x \right)} \right) = \left( {w + f'\left( {g\left( x \right)} \right)} \right)\left( {v + g'\left( x \right)} \right)h $$$$ {\left( {fog\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to o} \frac{{f\left( {g\left( {x + h} \right)} \right) - f\left( {g\left( x \right)} \right)}}{h} $$$$ {\left( {fog\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to o} \left( {w + f'\left( {g\left( x \right)} \right)} \right)\left( {v + g'\left( x \right)} \right) = f'\left( {g\left( x \right)} \right)g'\left( x \right) $$

Example 7: $ h(x) = (1+x^2)^{100}$ Justify $\left( {fog(x)} \right)' = g'(x)f'(g(x))$

Solution: Here, $$f(x) = x^{100}$$ $$g(x) = 1+x^2$$

We know that, $$\frac{d}{dx}x^n = nx^{n-1}$$ $$f'(x) = 100 x^{100-1} = 100 x^{99}$$

$$f'(g(x)) = f'(1+x^2) = 100 (1+x^2)^{99}$$ Using sum rule, $$g'(x) = 2x$$

Let $h(x) = fog(x)$,

$$h'(x) = {\left( {fog(x)} \right)^\prime } = 2x(100){(1 + {x^2})^{99}} = 200x{(1 + {x^2})^{99}}$$