\( \begin{array}{l}f(3)=2 f^{\prime}(3)=4 \\ \text { and } \\ g^{\prime}(3)=-3 g(3)=9\end{array} \rightarrow\left(\frac{f}{g}-f\right)^{\prime}(3)= \) ?

To find the derivative of the expression \(\left(\frac{f}{g}-f\right)\) at \(x=3\), we can use the quotient rule and the sum rule of differentiation. Given: \( f(3) = 2 \) \( f'(3) = 4 \) \( g'(3) = -3 \) \( g(3) = 9 \) Let's find the derivative of \(\frac{f}{g}\) at \(x=3\) first: \[ \left(\frac{f}{g}\right)'(3) = \frac{f'(3)g(3) - f(3)g'(3)}{(g(3))^2} \] Substitute the given values: \[ \left(\frac{f}{g}\right)'(3) = \frac{4 \cdot 9 - 2 \cdot (-3)}{9^2} = \frac{36 + 6}{81} = \frac{42}{81} \] Now, let's find the derivative of \(-f\) at \(x=3\): \[ (-f)'(3) = -f'(3) = -4 \] Finally, we can find the derivative of \(\left(\frac{f}{g}-f\right)\) at \(x=3\) by adding the derivatives of \(\frac{f}{g}\) and \(-f\): \[ \left(\frac{f}{g}-f\right)'(3) = \left(\frac{f}{g}\right)'(3) + (-f)'(3) = \frac{42}{81} - 4 \] Let's calculate the final result. Calculate the value by following steps: - step0: Calculate: \(\frac{42}{81}-4\) - step1: Reduce the fraction: \(\frac{14}{27}-4\) - step2: Reduce fractions to a common denominator: \(\frac{14}{27}-\frac{4\times 27}{27}\) - step3: Transform the expression: \(\frac{14-4\times 27}{27}\) - step4: Multiply the numbers: \(\frac{14-108}{27}\) - step5: Subtract the numbers: \(\frac{-94}{27}\) - step6: Rewrite the fraction: \(-\frac{94}{27}\) The derivative of \(\left(\frac{f}{g}-f\right)\) at \(x=3\) is \(-\frac{94}{27}\) or \(-3\frac{13}{27}\) or \(-3.4881\).