Questions: g(x) = x^2 / (1-x) then g'(x) equals

Transcript text: \[ g(x)=\frac{x^{2}}{1-x} \] then $g^{\prime}(x)$ equals

Solution

Solution Steps

To find the derivative $ g'(x) $ of the function $ g(x) = \frac{x^2}{1-x} $, we can use the quotient rule. The quotient rule states that if you have a function $ \frac{u(x)}{v(x)} $, its derivative is given by:

\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]

In this case, $ u(x) = x^2 $ and $ v(x) = 1-x $. We will find the derivatives $ u'(x) $ and $ v'(x) $, and then apply the quotient rule.

Step 1: Define the Function

We start with the function defined as: \[ g(x) = \frac{x^2}{1 - x} \]

Step 2: Apply the Quotient Rule

To find the derivative $ g'(x) $, we apply the quotient rule: \[ g'(x) = \frac{u'v - uv'}{v^2} \] where $ u = x^2 $ and $ v = 1 - x $.

Step 3: Calculate Derivatives

We calculate the derivatives:

$ u' = \frac{d}{dx}(x^2) = 2x $
$ v' = \frac{d}{dx}(1 - x) = -1 $

Step 4: Substitute into the Quotient Rule

Substituting $ u, u', v, $ and $ v' $ into the quotient rule gives: \[ g'(x) = \frac{(2x)(1 - x) - (x^2)(-1)}{(1 - x)^2} \]

Step 5: Simplify the Expression

Simplifying the numerator: \[ g'(x) = \frac{2x(1 - x) + x^2}{(1 - x)^2} = \frac{2x - 2x^2 + x^2}{(1 - x)^2} = \frac{2x - x^2}{(1 - x)^2} \]

Final Answer

Thus, the derivative of the function is: \[ \boxed{g'(x) = \frac{2x - x^2}{(1 - x)^2}} \]

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