Question

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

Answer 1

Given:$y=\frac{2x^9}{3}-\frac{5x^7}{7}+6x^3$

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

$=\frac{2}{3}\left(9x^8\right)-\frac{5}{7}\left(7x^6\right)+6\left(3x^2\right)$

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

$\therefore \frac{dy}{dx}=6x^8-5x^6+18x^2$ at $x=1$

$=6\times 1^8-5\times 1^6+18\times 1^2=6-5+18=19$