# Find the derivative of f(x) = \frac{\ln x}{\sqrt{1+x^2}} using the Quotient rule.

## Question:

Find the derivative of

{eq}f(x) = \frac{\ln x}{\sqrt{1+x^2}} {/eq}

using the Quotient rule.

## Using Quotient Rule for Derivatives:

The derivative of a function that is the quotient of two expressions is determined by the derivative of the numerator multiplied by the expression of the denominator minus the derivative of the denominator multiplied by the expression of the numerator, all divided by the expression of the square of the denominator: {eq}\,{\left[ {\frac{{u\left( x \right)}}{{v\left( x \right)}}} \right]^\prime } = \frac{{u^\prime{{\left( x \right)} } \cdot v\left( x \right) - u\left( x \right) \cdot {v^\prime }\left( x \right)}}{{{{\left( {v\left( x \right)} \right)}^2}}}{\text{.}} {/eq}

{eq}\eqalign{ & {\text{We're going to find the derivative for the function }}\,f\left( x \right) = \frac{{\ln x}}{{\sqrt {1 + {x^2}} }}.{\text{ }} \cr & {\text{Then}}{\text{, rewriting the expression:}} \cr & \,\,\,{f^\prime }\left( x \right) = {\left[ {\frac{{\ln x}}{{{{\left( {1 + {x^2}} \right)}^{\frac{1}{2}}}}}} \right]^\prime } \cr & {\text{Using the quotient rule }}\,{\left[ {\frac{{u\left( x \right)}}{{v\left( x \right)}}} \right]^\prime } = \frac{{u^\prime{{\left( x \right)} } \cdot v\left( x \right) - u\left( x \right) \cdot {v^\prime }\left( x \right)}}{{{{\left( {v\left( x \right)} \right)}^2}}}{\text{:}} \cr & \,\,\,{f^\prime }\left( x \right) = \frac{{{{\left( {\ln x} \right)}^\prime } \cdot {{\left( {1 + {x^2}} \right)}^{\frac{1}{2}}} - \ln x \cdot {{\left( {{{\left( {1 + {x^2}} \right)}^{\frac{1}{2}}}} \right)}^\prime }}}{{{{\left( {{{\left( {1 + {x^2}} \right)}^{\frac{1}{2}}}} \right)}^2}}} \cr & {\text{Applying the common derivative }}{\left( {\ln u} \right)^\prime } = \frac{1}{u}{u^\prime}{\text{ and the power rule }}{\left( {{u^n}} \right)^\prime } = n{u^{n - 1}}{u^\prime }{\text{:}} \cr & \,\,\,{f^\prime }\left( x \right) = \frac{{\frac{1}{x} \cdot {{\left( {1 + {x^2}} \right)}^{\frac{1}{2}}} - \ln x \cdot {{\left( {1 + {x^2}} \right)}^{ - \frac{1}{2}}}x}}{{1 + {x^2}}} \cr & \,\,\,{f^\prime }\left( x \right) = \frac{{\frac{{\sqrt {1 + {x^2}} }}{x} - \frac{{\ x ln x}}{{\sqrt {1 + {x^2}} }}}}{{1 + {x^2}}} \cr & {\text{Simplifying:}} \cr & \,\,\,{f^\prime }\left( x \right) = \frac{{\frac{{1 + {x^2} - {x^2}\ln x}}{{x\sqrt {1 + {x^2}} }}}}{{1 + {x^2}}} \cr & \,\,\,{f^\prime }\left( x \right) = \frac{{1 + {x^2} - {x^2}\ln x}}{{x\left( {1 + {x^2}} \right)\sqrt {1 + {x^2}} }} \cr & {\text{Therefore}}{\text{, the derivative of the function is: }}\boxed{\,{f^\prime }\left( x \right) = \frac{{1 + {x^2} - {x^2}\ln x}}{{x\left( {1 + {x^2}} \right)\sqrt {1 + {x^2}} }}} \cr} {/eq}