Differentiate (6x3 - 5x2 + 1) / 3x2 with respect to x.

Question

ScanSolve AI Tutor · Accepted Answer

To differentiate the given expression, we first simplify it by dividing each term in the numerator by the denominator.

Rewrite the expression by separating the terms.
$\frac{6x^3 - 5x^2 + 1}{3x^2} = \frac{6x^3}{3x^2} - \frac{5x^2}{3x^2} + \frac{1}{3x^2}$
Simplify each term using the rules of exponents.
$= 2x^{3-2} - \frac{5}{3}x^{2-2} + \frac{1}{3}x^{-2}$ $= 2x^1 - \frac{5}{3}x^0 + \frac{1}{3}x^{-2}$
Since $x^0 = 1$ , the expression becomes:
$= 2x - \frac{5}{3} + \frac{1}{3}x^{-2}$
Differentiate each term with respect to $x$ using the power rule $\frac{d}{dx}(ax^n) = anx^{n-1}$ and the rule that the derivative of a constant is zero.
$\frac{d}{dx}\left(2x - \frac{5}{3} + \frac{1}{3}x^{-2}\right) = \frac{d}{dx}(2x) - \frac{d}{dx}\left(\frac{5}{3}\right) + \frac{d}{dx}\left(\frac{1}{3}x^{-2}\right)$ $= 2(1)x^{1-1} - 0 + \frac{1}{3}(-2)x^{-2-1}$ $= 2x^0 - \frac{2}{3}x^{-3}$ $= 2(1) - \frac{2}{3}x^{-3}$ $= 2 - \frac{2}{3}x^{-3}$
Express the result with positive exponents.
$= 2 - \frac{2}{3x^3}$
The final answer is $2 - \frac{2}{3x^3}$ .

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