QuestionJuly 20, 2025

Find the derivative of: h(x)=(x^3+3x+4)/(x^3)+3x-2

Find the derivative of: h(x)=(x^3+3x+4)/(x^3)+3x-2
Find the derivative of:
h(x)=(x^3+3x+4)/(x^3)+3x-2

Solution
4.3(231 votes)

Answer

0 Explanation 1. Identify the Quotient Rule Use the quotient rule for derivatives: **\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}**, where u = x^3 + 3x + 4 and v = x^3 + 3x - 2. 2. Differentiate the Numerator u' = \frac{d}{dx}(x^3 + 3x + 4) = 3x^2 + 3. 3. Differentiate the Denominator v' = \frac{d}{dx}(x^3 + 3x - 2) = 3x^2 + 3. 4. Apply the Quotient Rule Substitute u, v, u', and v' into the quotient rule formula: \frac{(3x^2 + 3)(x^3 + 3x - 2) - (x^3 + 3x + 4)(3x^2 + 3)}{(x^3 + 3x - 2)^2}. 5. Simplify the Expression Expand and simplify the numerator: (3x^2 + 3)(x^3 + 3x - 2) - (x^3 + 3x + 4)(3x^2 + 3). This simplifies to 0 after expanding both products and subtracting.

Explanation

1. Identify the Quotient Rule<br /> Use the quotient rule for derivatives: **$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$**, where $u = x^3 + 3x + 4$ and $v = x^3 + 3x - 2$.<br /><br />2. Differentiate the Numerator<br /> $u' = \frac{d}{dx}(x^3 + 3x + 4) = 3x^2 + 3$.<br /><br />3. Differentiate the Denominator<br /> $v' = \frac{d}{dx}(x^3 + 3x - 2) = 3x^2 + 3$.<br /><br />4. Apply the Quotient Rule<br /> Substitute $u$, $v$, $u'$, and $v'$ into the quotient rule formula: <br /> $\frac{(3x^2 + 3)(x^3 + 3x - 2) - (x^3 + 3x + 4)(3x^2 + 3)}{(x^3 + 3x - 2)^2}$.<br /><br />5. Simplify the Expression<br /> Expand and simplify the numerator: $(3x^2 + 3)(x^3 + 3x - 2) - (x^3 + 3x + 4)(3x^2 + 3)$.<br /> This simplifies to $0$ after expanding both products and subtracting.
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