Finding an Equation of a Tangent Line

Find an equation of the tangent line to the graph of \(f( x) =\dfrac{x\sqrt{x^{2}+3}}{1+x}\) at the point \((1,1) .\)

Solution The slope of the tangent line to the graph of \(y=f( x) \) at the point \((1,1) \) is \(f^\prime (1) \). Since the function consists of a product, a quotient, and a power, we follow the steps for logarithmic differentiation:

Step 1 Take the natural logarithm of each side, and simplify: \[ \ln y=\ln \dfrac{x\sqrt{x^{2}+3}}{1+x}=\ln x+\dfrac{1}{2}\ln ( x^{2}+3) -\ln ( 1+x) \]

Step 2 Differentiate implicitly. \[ \dfrac{y^\prime }{y}=\dfrac{1}{x}+\dfrac{1}{2}\cdot \dfrac{2x}{x^{2}+3}- \dfrac{1}{1+x}=\dfrac{1}{x}+\dfrac{x}{x^{2}+3}-\dfrac{1}{1+x} \]

Step 3 Solve for \(y^\prime \) and simplify: \[ y^\prime =f^\prime (x) = y\left( \dfrac{1}{x}+\dfrac{x}{x^{2}+3}-\dfrac{1}{1+x}\right) = \dfrac{x\sqrt{x^{2}+3}}{1+x}\left( \dfrac{1}{x}+\dfrac{x}{x^{2}+3}-\dfrac{1}{1+x}\right) \]

Now we find the slope of the tangent line by evaluating \(f^\prime (1) \). \[ f^\prime (1) =\dfrac{\sqrt{4}}{2}\left( 1+\dfrac{1}{4}-\dfrac{1}{2}\right) =\dfrac{3}{4} \]

Then an equation of the tangent line to the graph of \(f\) at the point \((1,1) \) is \[ \begin{eqnarray*} y-1 &=&\dfrac{3}{4}( x-1) \\ y &=&\dfrac{3}{4}x+\dfrac{1}{4} \end{eqnarray*} \]