User:Arpan Mathur/sandbox

Differentiation
Example 1

The logarithmic function f(x) = logex is differentiable at x = 5 and its derivative there is 1/5.

This result may also be established by calculating the limit as h approaches zero of the d.c.(difference quotient) of f(5).

$$ f'(5) = \lim_{h \to 0}\frac{f(5+h)-f(5)}{h} = \lim_{h \to 0}\frac{\log_e(5+h)-\log_e(5)}{h} = \lim_{h \to 0}\frac{1}{h}\log_e\left(\frac{5+h}{5}\right)$$

$$ = \lim_{h \to 0}\frac{1}{h} \log_e\left(1 + \frac{h}{5}\right) $$

Now ,the series expansion of $$ \log_e(1+x) $$ is :

$$ \log_e(1+x) = x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7 + ........ $$

Therefore

$$ \log_e\left(1+\frac{h}{5} \right) = \frac{h}{5} - \frac{1}{3}\left(\frac{h}{5}\right)^3 + \frac{1}{5}\left(\frac{h}{5}\right)^5 - \frac{1}{7}\left(\frac{h}{5}\right)^7 + ....... $$

and the derivative is therefore

$$ f'(5) = \lim_{h \to 0} \left( \frac{1}{5} - \frac{h^2}{3 \times 5^3} + \frac{h^4}{5 \times 5^5} -\frac{h^6}{7 \times 5^7} + ....... \right) = \frac{1}{5}$$

Hence the slope of the graph of the logarithmic function at the point (5, loge 5) is 1/5, and so its derivative at x = 5 is f′(5) = 1/5.

A more general computation shows that the derivative of the logarithmic function f(x) = logex at x = a is f'(a) = 1/a.

Integration
Example 2

\begin{align} \int \frac{x^2 + 5}{(x + 1)(x + 2)(x + 3)} \;\mathrm{d}x & = \int \left(\frac{3}{x + 1}- \frac{9}{x + 2}+\frac{7}{x + 3} \right) \;\mathrm{d}x\\ & = 3\ln|x + 1| - 9\ln|x + 2| + 7\ln|x + 3| + C    \\ \end{align} $$ Example 3

\begin{align} & \int \log x \;\mathrm{d}x = \int (\log x . 1)\;\mathrm{d}x = \log x \int 1 \;\mathrm{d}x - \int \left( \frac{\;\mathrm{d}}{\;\mathrm{d}x}(\log x) \int 1 \;\mathrm{d}x \right)\;\mathrm{d}x \\ & = x\log x - \int \left( \frac{1}{x}. x \right) \;\mathrm{d}x = x \log x - \int 1 \;\mathrm{d}x = x\log x - x + C \\ & = x(\log x - 1) + C\\ \end{align} $$