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Problem 971

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

Expert verified
The derivative of $$g(x) = \frac{x^3}{2\sqrt{3x^2 - 1}}$$ is $$g'(x) = \frac{3x^2 - 2x^3}{4(3x^2 - 1)}$$.
See the step by step solution

## 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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