Questions: Differentiate. y=ln(5x^2-3x+4) y'=

Transcript text: Differentiate. \[ \begin{array}{l} y=\ln \left(5 x^{2}-3 x+4\right) \\ y^{\prime}=\square \end{array} \]

Solution

Solution Steps

Step 1: Define the Function

We start with the function given by the logarithmic expression: $y = \ln(5x^2 - 3x + 4)$

Step 2: Differentiate the Function

To find the derivative $y'$ , we apply the chain rule. The derivative of $\ln(u)$ is $\frac{1}{u} \cdot \frac{du}{dx}$ , where $u = 5x^2 - 3x + 4$ . Thus, we have: $y' = \frac{1}{5x^2 - 3x + 4} \cdot (10x - 3)$

Step 3: Simplify the Derivative

Combining the results, we can express the derivative as: $y' = \frac{10x - 3}{5x^2 - 3x + 4}$

Final Answer

The derivative of the function is: $\boxed{y' = \frac{10x - 3}{5x^2 - 3x + 4}}$

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