find ( f(3) ).\n( f(3)=square )\nuse this to find an equation of the tangent line to the curve ( y = f(x) )…

Explanation:

Step1: Differentiate ( f(x)=\frac{3x}{1 + x^{2}} )

Use the quotient rule ( \left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}} ), where ( u = 3x ), ( u'=3 ), ( v = 1 + x^{2} ), ( v' = 2x ). [ \begin{align*} f'(x)&=\frac{3(1 + x^{2})-3x(2x)}{(1 + x^{2})^{2}}\ &=\frac{3+3x^{2}-6x^{2}}{(1 + x^{2})^{2}}\ &=\frac{3 - 3x^{2}}{(1 + x^{2})^{2}} \end{align*} ]

Step2: Find ( f'(3) )

Substitute ( x = 3 ) into ( f'(x) ). [ \begin{align*} f'(3)&=\frac{3-3\times3^{2}}{(1 + 3^{2})^{2}}\ &=\frac{3-27}{(1 + 9)^{2}}\ &=\frac{-24}{100}\ &=-\frac{6}{25} \end{align*} ]

Step3: Find the equation of the tangent line

The point - slope form of a line is ( y - y_{0}=m(x - x_{0}) ), where ( (x_{0},y_{0})=(3,\frac{9}{10}) ) and ( m = f'(3)=-\frac{6}{25} ). [ \begin{align*} y-\frac{9}{10}&=-\frac{6}{25}(x - 3)\ y-\frac{9}{10}&=-\frac{6}{25}x+\frac{18}{25}\ y&=-\frac{6}{25}x+\frac{18}{25}+\frac{9}{10}\ y&=-\frac{6}{25}x+\frac{36 + 45}{50}\ y&=-\frac{6}{25}x+\frac{81}{50} \end{align*} ]

Answer:

( f'(3)=-\frac{6}{25} ), the equation of the tangent line is ( y =-\frac{6}{25}x+\frac{81}{50} )