If $f ( x ) = \frac { x ^ { 2 } - 3 x + 5 } { \sqrt { 1 - 3 x } }$ , then $f ^ { \prime } ( x ) = \frac { 2 x - 3 } { \sqrt { 1 - 3 x } }$ .

Question

Quizplus · Accepted Answer

The derivative $f'(x)$ of the given function $f(x) = \frac{x^2 - 3x + 5}{\sqrt{1 - 3x}}$ involves applying the quotient rule and the chain rule, which would result in a more complex expression than $\frac{2x - 3}{\sqrt{1 - 3x}}$. The derivative of the numerator and the adjustment for the derivative of the denominator (due to the chain rule) are not accounted for in the provided expression.

If f(x)=x2−3x+51−3xf ( x ) = \frac { x ^ { 2 } - 3 x + 5 } { \sqrt { 1 - 3 x } }f(x)=1−3x​x2−3x+5​

If $f ( x ) = \frac { x ^ { 2 } - 3 x + 5 } { \sqrt { 1 - 3 x } }$