Question

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

Answer 1

We have,

$\frac{dy}{dx}=(\frac{2x^9}{3}-\frac{5}{7}{x}^7+6x^3-x)$

$=\frac{2}{3}\frac{d{x}^9}{dx}-\frac{5}{7}\frac{d{x}^7}{dx}+6\frac{d{x}^3}{dx}-\frac{dx}{dx}$

$=\frac{2}{3}.9x^8-\frac{5}{7}.7x^6+18x^2-1$

$=6x^8-5x^6+18x^2-1$

$\therefore \frac{dy}{dx}$ at $x=1$

$=6(1{)}^8-5(1{)}^6+18(1{)}^2-1$

$=6-5+18-1$

=18