3.3 Quotient 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

$$\frac{d}{{dx}}\left( {\frac{{f(x)}}{{g(x)}}} \right) = \frac{{g(x)\frac{d}{{dx}}f(x) - f(x)\frac{d}{{dx}}g(x)}}{{{{\left( {g(x)} \right)}^2}}}$$ Proof:

$$ {\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{f\left( {x + h} \right)}}{{g\left( {x + h} \right)}}} \right) + \left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)}}{h} $$

$$ {\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\left( {\frac{{f\left( {x + h} \right)}}{{g\left( {x + h} \right)}}} \right) + \left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)} \right) $$

$$ {\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{{f\left( {x + h} \right)g\left( x \right) + f\left( x \right)g\left( {x + h} \right)}}{{g\left( {x + h} \right)g\left( x \right)}}} \right) $$

Add and subtract $f\left( x \right)g\left( x \right)$ in numerator,

$$ {\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{{\left( {f\left( {x + h} \right)g\left( x \right)} \right) - f\left( x \right)g\left( x \right)\, + f\left( x \right)g\left( x \right)\, - \left( {f\left( x \right)g\left( {x + h} \right)} \right)}}{{g\left( {x + h} \right)g\left( x \right)}}} \right) $$

$$ {\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left( {\frac{{\left( {f\left( {x + h} \right) - f\left( x \right)} \right)g\left( x \right)\, + f\left( x \right)\,\left( {f\left( x \right) - f\left( {x + h} \right)} \right)}}{{g\left( {x + h} \right)g\left( x \right)}}} \right) $$

$$ {\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{1}{{g\left( {x + h} \right)g\left( x \right)}}\left( {\frac{{\left( {f\left( {x + h} \right) - f\left( x \right)} \right)g\left( x \right)\, - f\left( x \right)\,\left( {f\left( x \right) + f\left( {x + h} \right)} \right)}}{h}} \right) $$

$$ {\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{1}{{g\left( {x + h} \right)g\left( x \right)}}\left( {\frac{{\left( {f\left( {x + h} \right) - f\left( x \right)} \right)g\left( x \right)\,}}{h} - \frac{{f\left( x \right)\,\left( {f\left( x \right) + f\left( {x + h} \right)} \right)}}{h}} \right) $$

$$ {\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{1}{{g\left( {x + h} \right)g\left( x \right)}}\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\left( {f\left( {x + h} \right) - f\left( x \right)} \right)g\left( x \right)\,}}{h} - \mathop {\lim }\limits_{h \to 0} \frac{{f\left( x \right)\,\left( {f\left( x \right) + f\left( {x + h} \right)} \right)}}{h}} \right) $$

$$ {\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{1}{{g\left( {x + h} \right)g\left( x \right)}}\left( {\mathop {\lim }\limits_{h \to 0} g\left( x \right)\mathop {\lim }\limits_{h \to 0} \frac{{\left( {f\left( {x + h} \right) - f\left( x \right)} \right)\,}}{h} - \mathop {\lim }\limits_{h \to 0} f\left( x \right)\mathop {\lim }\limits_{h \to 0} \frac{{\,\left( {f\left( x \right) + f\left( {x + h} \right)} \right)}}{h}} \right) $$

$$ {\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)^\prime } = \frac{1}{{g\left( x \right)g\left( x \right)}}\left( {g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)} \right) $$

$$ {\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)^\prime } = \frac{{\left( {g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} $$

Example 6: $f(x)= x^4 $ and $g(x) = x^8 $ prove $\frac{d}{{dx}}\left( {\frac{{f(x)}}{{g(x)}}} \right)= \frac{{g(x)\frac{d}{{dx}}f(x) - f(x)\frac{d}{{dx}}g(x)}}{{{{\left( {g(x)} \right)}^2}}}$

Solution: Given: $$f(x)= x^4 $$ $$g(x) = x^8 $$ $$\left( {\frac{{f(x)}}{{g(x)}}} \right) = \frac{x^4}{x^8} = \frac{1}{x^4}= x^{-4}$$

L.H.S: $$\frac{d}{{dx}}\left( {\frac{{f(x)}}{{g(x)}}} \right) = \frac{d}{{dx}} x^{-4}$$ We know that, $$\frac{d}{dx}x^n = nx^{n-1}$$

$$\frac{d}{{dx}} x^{-4} = -4 x^{-4-1} = -4x^{-5} = \frac{-4}{x^5}$$

R.H.S: $$\frac{{g(x)\frac{d}{{dx}}f(x) - f(x)\frac{d}{{dx}}g(x)}}{{{{\left( {g(x)} \right)}^2}}} = \frac{{{x^8}\frac{d}{{dx}}{x^4} -{x^4}\frac{d}{{dx}}{x^8}}}{{{{\left( {{x^8}} \right)}^2}}}$$ we know that $$\frac{d}{dx}x^n = nx^{n-1}$$ $$\frac{{{x^8}\frac{d}{{dx}}{x^4} - {x^4}\frac{d}{{dx}}{x^8}}}{{{{\left( {{x^8}} \right)}^2}}} = \frac{{{x^8}(4{x^{4 - 1}}) - {x^4}(8{x^{8 - 1}})}}{{\left( {{x^{16}}} \right)}}$$ $$ \frac{{{x^8}\frac{d}{{dx}}{x^4} - {x^4}\frac{d}{{dx}}{x^8}}}{{{{\left( {{x^8}} \right)}^2}}} = \frac{{{x^8}(4{x^3}) - {x^4}(8{x^7})}}{{\left( {{x^{16}}} \right)}}$$ $$ \frac{{{x^8}\frac{d}{{dx}}{x^4} - {x^4}\frac{d}{{dx}}{x^8}}}{{{{\left( {{x^8}} \right)}^2}}} = \frac{{4{x^{11}} - 8{x^{11}}}}{{\left( {{x^{16}}} \right)}}$$ $$ \frac{{{x^8}\frac{d}{{dx}}{x^4} - {x^4}\frac{d}{{dx}}{x^8}}}{{{{\left( {{x^8}} \right)}^2}}} = \frac{{- 4{x^{11}}}}{{\left( {{x^{16}}} \right)}}$$ $$ \frac{{{x^8}\frac{d}{{dx}}{x^4} - {x^4}\frac{d}{{dx}}{x^8}}}{{{{\left( {{x^8}} \right)}^2}}} = \frac{{- 4}}{{\left( {{x^{5}}} \right)}}$$

Thus, L.H.S = R.H.S