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Problem 971

Find the derivative of: $\left.g(x)=x^{3} /\left[{ }^{2} \sqrt{(} 3 x^{2}-1\right)\right]$.

Short Answer

Expert verified
The derivative of \(g(x) = \frac{x^3}{2\sqrt{3x^2 - 1}}\) is \(g'(x) = \frac{3x^2 - 2x^3}{4(3x^2 - 1)}\).
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Step 1: Identify the outside and inside function

In the given function, \( g(x) = \frac{x^3}{2 \sqrt{3x^2 - 1}} \), the outside function is the division operation of \( x^3 \) by the square root part, and the inside function is \( 3x^2 - 1 \) under the square root.

Step 2: Apply the Quotient Rule

The quotient rule states that the derivative of \(\frac{u}{v}\) is \(\frac{vu' - uv'}{v^2}\), where \(u'\) is the derivative of u and \(v'\) is the derivative of v. Here, \(u = x^3\) and \(v = 2\sqrt{3x^2 - 1}\). First, find the derivatives of \(u\) and \(v\): - Derivative of \(u = x^3\) is \(u' = 3x^2\) using the power rule. - Derivative of \(v = 2\sqrt{3x^2 - 1}\) requires the chain rule since \(3x^2 - 1\) is nested inside the square root function.

Step 3: Apply the Chain Rule for \(v'\)

The derivative of the function \(v = 2\sqrt{3x^2 - 1}\) is \(v' = 2 \frac{1}{2\sqrt{3x^2 - 1}} \cdot (6x)\). The \(2\) and \(\frac{1}{2}\) cancel out, and the slope is \(v' = \frac{6x}{\sqrt{3x^2 - 1}}\).

Step 4: Substitute Expressions in Quotient Rule

Substitute \(u', v'\) and \(v\) into the quotient rule formula. Thus, the derivative of the function \(g = \frac{u'}{v} = \frac{v(3x^2) - u * \frac{6x}{\sqrt{3x^2 - 1}}}{(2 \sqrt{3x^2 - 1})^2}\).

Step 5: Simplify

The function can be further simplified: - Multiply \(v\) and \(3x^2\) to get \(6x^5 - 2x^3\) - The denominator when squared becomes \(4(3x^2-1)\) Therefore, the derivative of \( \frac{x^3}{2 \sqrt{3x^2 - 1}}\) is \( \frac{3x^2 * 2\sqrt{3x^2 - 1} - x^3 * \frac{6x}{\sqrt{3x^2 - 1}}}{4(3x^2 - 1)}\). After some simplification, you get \(g'(x) = \frac{3x^2 - 2x^3}{4(3x^2 -1)}\).

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