{:[" 2.) "g(x)=(sqrt(4x^(3)-3x^(2)+2x-1))^(3)],[g^(')(x)=((4x^(3)-3x^(2)+2x-1)^(1//2))^(3)]:}

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Apply Chain Rule: To find the derivative of the function $ g(x) = (\sqrt{4x^3 - 3x^2 + 2x - 1})^3 $, we will use the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.Identify Functions: First, let's identify the outer function and the inner function. The outer function is $ u^3 $ where $ u $ is the inner function. The inner function is $ \sqrt{4x^3 - 3x^2 + 2x - 1} $, which can also be written as $ (4x^3 - 3x^2 + 2x - 1)^{1/2} $.Derivative of Outer Function: The derivative of the outer function $ u^3 $ with respect to $ u $ is $ 3u^2 $. We will later substitute $ u $ with the inner function.Derivative of Inner Function: Now, we need to find the derivative of the inner function $ (4x^3 - 3x^2 + 2x - 1)^{1/2} $. To do this, we use the power rule combined with the chain rule. The derivative of $ u^{1/2} $ with respect to $ u $ is $ \frac{1}{2}u^{-1/2} $.Combine Derivatives: Next, we need to find the derivative of $ 4x^3 - 3x^2 + 2x - 1 $ with respect to $ x $. Using the power rule, the derivative is $ 12x^2 - 6x + 2 $.Final Derivative: Now we can combine all the parts. The derivative of the inner function with respect to $ x $ is $ \frac{1}{2}(4x^3 - 3x^2 + 2x - 1)^{-1/2} \cdot (12x^2 - 6x + 2) $.Simplify Expression: Finally, we multiply the derivative of the outer function by the derivative of the inner function to get the derivative of the composite function: $$ g'(x) = 3(\sqrt{4x^3 - 3x^2 + 2x - 1})^2 \cdot \frac{1}{2}(4x^3 - 3x^2 + 2x - 1)^{-1/2} \cdot (12x^2 - 6x + 2) $$Simplify the expression by canceling out the $ (4x^3 - 3x^2 + 2x - 1)^{-1/2} $ term in the derivative of the inner function with one of the $ (4x^3 - 3x^2 + 2x - 1)^{1/2} $ terms in the outer function's derivative: $$ g'(x) = 3 \cdot \frac{1}{2} \cdot (12x^2 - 6x + 2) $$ $$ g'(x) = \frac{3}{2} \cdot (12x^2 - 6x + 2) $$ $$ g'(x) = 18x^2 - 9x + 3 $$

$\begin{array}{l}\text { 2.) } g(x)=\left(\sqrt{4 x^{3}-3 x^{2}+2 x-1}\right)^{3} \\ g^{\prime}(x)=\left(\left(4 x^{3}-3 x^{2}+2 x-1\right)^{1 / 2}\right)^{3}\end{array}$

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2.) g(x)=(4x3−3x2+2x−1)3g′(x)=((4x3−3x2+2x−1)1/2)3 \begin{array}{l}\text { 2.) } g(x)=\left(\sqrt{4 x^{3}-3 x^{2}+2 x-1}\right)^{3} \\ g^{\prime}(x)=\left(\left(4 x^{3}-3 x^{2}+2 x-1\right)^{1 / 2}\right)^{3}\end{array} 2.) g(x)=(4x3−3x2+2x−1​)3g′(x)=((4x3−3x2+2x−1)1/2)3​

$\begin{array}{l}\text { 2.) } g(x)=\left(\sqrt{4 x^{3}-3 x^{2}+2 x-1}\right)^{3} \\ g^{\prime}(x)=\left(\left(4 x^{3}-3 x^{2}+2 x-1\right)^{1 / 2}\right)^{3}\end{array}$