Question

$\mathrm{If}y=\frac{2x^9}{3}-\frac{5}{7}{x}^7+6x^3-x,\mathrm{find}\frac{dy}{dx}\mathrm{at}x=1.$

Answer 1

$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(\frac{2x^9}{3}-\frac{5}{7}{x}^7+6x^3-x\right) \\ =\frac{2}{3}\frac{d}{dx}\left(x^9\right)-\frac{5}{7}\frac{d}{dx}\left(x^7\right)+6\frac{d}{dx}\left(x^3\right)-\frac{d}{dx}\left(x\right) \\ =\frac{2}{3}\left(9x^8\right)-\frac{5}{7}\left(7x^6\right)+6\left(3x^2\right)-1 \\ =6x^8-5x^6+18x^2-1 \\ \frac{dy}{dx}\text{at x = 1:} \\ 6{\left(1\right)}^8-5{\left(1\right)}^6+18{\left(1\right)}^2-1 \\ =6-5+18-1 \\ =18 \end{gathered}$$