If y= 2x^9 3 - 5x^7 7 +6x^3-x, find dy dx at x=1 - Vidyadip

Question

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

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Answer

We have,$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{2x}^9}{3}-\frac{\text{5x}^7}{7}+\text{6x}^3-\text{x}\Big)$
$=\frac{2}{3}\frac{\text{d}}{\text{dx}}(\text{x})^9-\frac{5}{7}\frac{\text{d}}{\text{dx}}(\text{x})^7+6\frac{\text{d}}{\text{dx}}(\text{x})^3-\frac{\text{x}}{\text{dx}}(\text{x})$
$=\frac{2}{3}.9\text{x}^8-\frac{5}{7}.7\text{x}^6+\text{18x}^2-1$
$=\text{6x}^8-\text{5x}^6+\text{18x}^2-1$
$\therefore\frac{\text{dy}}{\text{dx}}$ at $\text{x}=1$
$=\text{6}(1)^8-\text{5}(1)^6+\text{18}(1)^2-1$
$=6-5+18-1$
$=18$

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