How do you find the derivative of $\frac{6 x^{5} + 3 x^{3} + x}{x^{2}}$ $(6x^5 + 3x^3 +x)/x^2$ ?

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How do you find the derivative of $\frac{6 x^{5} + 3 x^{3} + x}{x^{2}}$ $(6x^5 + 3x^3 +x)/x^2$ ?

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Answer 1 · 2015-10-24T14:34:25

$\frac{d}{d x} \frac{6 x^{5} + 3 x^{3} + x}{x^{2}} = \frac{(x^{2}) (30 x^{4} + 9 x^{2} + 1) - 2 x (6 x^{5} + 3 x^{3} + x)}{x^{4}}$ $d/dx(6x^5+3x^3+x)/x^2=((x^2)(30x^4+9x^2+1)-2x(6x^5+3x^3+x))/x^4$

Explanation:

We may use the quotient rule to differentiate this function to obtain :

$\frac{d}{d x} \frac{6 x^{5} + 3 x^{3} + x}{x^{2}} = \frac{(x^{2}) (30 x^{4} + 9 x^{2} + 1) - 2 x (6 x^{5} + 3 x^{3} + x)}{x^{4}}$ $d/dx(6x^5+3x^3+x)/x^2=((x^2)(30x^4+9x^2+1)-2x(6x^5+3x^3+x))/x^4$

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How do you find the derivative of $\frac{6 x^{5} + 3 x^{3} + x}{x^{2}}$ $(6x^5 + 3x^3 +x)/x^2$ ?

1 Answer

Explanation:

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How do you find the derivative of $\frac{6 x^{5} + 3 x^{3} + x}{x^{2}}$ $(6x^5 + 3x^3 +x)/x^2$ ?