QuestionJuly 23, 2025

Compute the derivative of the function f(x)=((3x^2-x+1)/(x^2)+1)^7 Select one: a 7((3x^2-x+1)/(x^2)+1)^6 b. 7((3x^2-x+1)/(x^2)+1)^6cdot (x^2+4x-1)/((x^2)+1)^(2) C. 7((3x^2-x+1)/(x^2)+1)^6cdot (1)/((x^2)+1)^(2) d. ((3x^2-x+1)/(x^2)+1)^6cdot (x^2+4x-1)/((x^2)+1)^(2)

Compute the derivative of the function f(x)=((3x^2-x+1)/(x^2)+1)^7 Select one: a 7((3x^2-x+1)/(x^2)+1)^6 b. 7((3x^2-x+1)/(x^2)+1)^6cdot (x^2+4x-1)/((x^2)+1)^(2) C. 7((3x^2-x+1)/(x^2)+1)^6cdot (1)/((x^2)+1)^(2) d. ((3x^2-x+1)/(x^2)+1)^6cdot (x^2+4x-1)/((x^2)+1)^(2)
Compute the derivative of the function f(x)=((3x^2-x+1)/(x^2)+1)^7 Select one: a 7((3x^2-x+1)/(x^2)+1)^6 b. 7((3x^2-x+1)/(x^2)+1)^6cdot (x^2+4x-1)/((x^2)+1)^(2) C. 7((3x^2-x+1)/(x^2)+1)^6cdot (1)/((x^2)+1)^(2) d. ((3x^2-x+1)/(x^2)+1)^6cdot (x^2+4x-1)/((x^2)+1)^(2)

Solution
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Answer

b. 7\left(\frac{3x^2 - x + 1}{x^2 + 1}\right)^6 \cdot \frac{x^2 + 4x - 1}{(x^2 + 1)^2} Explanation 1. Apply the Chain Rule Differentiate f(x) = \left(\frac{3x^2 - x + 1}{x^2 + 1}\right)^7 using the chain rule. The derivative is 7\left(\frac{3x^2 - x + 1}{x^2 + 1}\right)^6 \cdot \frac{d}{dx}\left(\frac{3x^2 - x + 1}{x^2 + 1}\right). 2. Differentiate the Quotient Use the quotient rule for \frac{d}{dx}\left(\frac{3x^2 - x + 1}{x^2 + 1}\right). **Quotient Rule**: \frac{u}{v} gives \frac{u'v - uv'}{v^2}. Here, u = 3x^2 - x + 1, v = x^2 + 1. - u' = 6x - 1 - v' = 2x 3. Apply the Quotient Rule Substitute into the quotient rule: \frac{(6x - 1)(x^2 + 1) - (3x^2 - x + 1)(2x)}{(x^2 + 1)^2}. 4. Simplify the Derivative Simplify the numerator: (6x - 1)(x^2 + 1) - (3x^2 - x + 1)(2x) = 6x^3 + 6x - x^2 - 1 - 6x^3 + 2x^2 - 2x = x^2 + 4x - 1. 5. Combine Results Combine with the chain rule result: 7\left(\frac{3x^2 - x + 1}{x^2 + 1}\right)^6 \cdot \frac{x^2 + 4x - 1}{(x^2 + 1)^2}.

Explanation

1. Apply the Chain Rule<br /> Differentiate $f(x) = \left(\frac{3x^2 - x + 1}{x^2 + 1}\right)^7$ using the chain rule. The derivative is $7\left(\frac{3x^2 - x + 1}{x^2 + 1}\right)^6 \cdot \frac{d}{dx}\left(\frac{3x^2 - x + 1}{x^2 + 1}\right)$.<br /><br />2. Differentiate the Quotient<br /> Use the quotient rule for $\frac{d}{dx}\left(\frac{3x^2 - x + 1}{x^2 + 1}\right)$. **Quotient Rule**: $\frac{u}{v}$ gives $\frac{u'v - uv'}{v^2}$. Here, $u = 3x^2 - x + 1$, $v = x^2 + 1$.<br />- $u' = 6x - 1$<br />- $v' = 2x$<br /><br />3. Apply the Quotient Rule<br /> Substitute into the quotient rule: $\frac{(6x - 1)(x^2 + 1) - (3x^2 - x + 1)(2x)}{(x^2 + 1)^2}$.<br /><br />4. Simplify the Derivative<br /> Simplify the numerator: $(6x - 1)(x^2 + 1) - (3x^2 - x + 1)(2x) = 6x^3 + 6x - x^2 - 1 - 6x^3 + 2x^2 - 2x = x^2 + 4x - 1$.<br /><br />5. Combine Results<br /> Combine with the chain rule result: $7\left(\frac{3x^2 - x + 1}{x^2 + 1}\right)^6 \cdot \frac{x^2 + 4x - 1}{(x^2 + 1)^2}$.
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