Answer to Question #134604 in Calculus for Ojugbele Daniel

Question

Differentiate x^20 +2x -10 from first principle

Answer 1

$\textsf{Let}\hspace{0.1cm}\displaystyle f(x) = x^{20} + 2x - 10\\ f(x + \delta x) = \\(x + \delta x)^{20} + 2(x + \delta x) - 10=\\ \sum_{r = 0}^{20} \binom{20}{r} x^{20-r} (\delta x)^r + 2x + 2\delta x - 10\hspace{1cm} \textsf{(Binomial theorem)}\\ f(x + \delta x) - f(x) = \sum_{r = 0}^{20} \binom{20}{r}x^{20-r} (\delta x)^r + 2x + 2\delta x - 10 - (x^{20} + 2x - 10)= \\ \sum_{r = 1}^{20}\binom{20}{r} x^{20-r} (\delta x)^r + x^{20} + 2x + 2\delta x - 10 - (x^{20} + 2x - 10) = \\ \sum_{r = 1}^{20} \binom{20}{r}x^{20-r} (\delta x)^r + 2\delta x \\ \frac{f(x + \delta x) - f(x)}{\delta x} = \sum_{r = 1}^{20}\binom{20}{r}x^{20-r} (\delta x)^{r - 1} + 2 = \binom{20}{1}x^{19} + \sum_{r = 2}^{20}\binom{20}{r}x^{20-r} (\delta x)^{r - 1} + 2 =\\20x^{19} + \sum_{r = 2}^{20}\binom{20}{r} x^{20-r} (\delta x)^{r - 1} + 2\\ \Rightarrow\lim_{\delta x \rightarrow 0}\frac{f(x + \delta x) - f(x)}{\delta x} =\\ f'(x) = 20x^{19} + 2 + \lim_{\delta x \rightarrow 0} \left(\binom{20}{2} x^{18} \delta x +...+(\delta x)^{19}\right) = 20x^{19} + 2\\ \therefore f'(x) = 20x^{19} + 2$