Question

Differentiate the function.

3. \(f\left( x \right) = ln\left( {{x^2} + 3x + 5} \right)\)

Short answer

The derivative of the function is \(f'\left( x \right) = \frac{{2x + 3}}{{{x^2} + 3x + 5}}\).

Step-by-step solution

Use the derivative of logarithmic function

Rule 2: The derivative of \(\ln x\) is,

\(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\)

Evaluating the derivative of given function

\(f\left( x \right) = \ln \left( {{x^2} + 3x + 5} \right)\)

Differentiate \(f\left( x \right)\)with respect to x

\(\begin{aligned}{l}\frac{{df\left( x \right)}}{{dx}} &= \frac{{d\log \left( {{x^2} + 3x + 5} \right)}}{{d\left( {{x^2} + 3x + 5} \right)}} \times \frac{{d\left( {{x^2} + 3x + 5} \right)}}{{dx}}\\f'\left( x \right) &= \frac{1}{{\left( {{x^2} + 3x + 5} \right)}} \times \left( {\frac{{d{x^2}}}{{dx}} + \frac{{d3x}}{{dx}} + \frac{{d5}}{{dx}}} \right)\\f'\left( x \right) &= \frac{1}{{\left( {{x^2} + 3x + 5} \right)}}\left( {2x + 3 + 0} \right)\\f'\left( x \right) &= \frac{{2x + 3}}{{{x^2} + 3x + 5}}\end{aligned}\)

Thus, the value of the derivative is \(f'\left( x \right) = \frac{{2x + 3}}{{{x^2} + 3x + 5}}\).