Consider the function $y(x) =\ln(x^2 + 1)$. Determine the slope of the tangent line to the graph of $y(x)$ in $(3, \ln(10))$.
This slope is equal to the derivative of $y(x)$ in $x=3$. To determine $y'(x)$ we need the chain rule and therefore we have to write $y(x)$ as $u(v(x))$. We choose $v(x) = x^2+1$ and $u(v) = \ln(v)$. Now we can determine $y'(x)$ and consequently, $y'(3)$:
$$
\begin{align}
v'(x) &= 2x + 0 = 2x\\
u'(v) &= \dfrac{1}{v}\\
y'(x) &= u'\big(v(x)\big)\cdot v'(x) = \dfrac{1}{v(x)} \cdot 2x = \dfrac{2x}{x^2+1}\\
y'(3) &= \dfrac{2\cdot 3}{3^2+1} = \dfrac{6}{10} = \dfrac{3}{5}.
\end{align}
$$
Hence, the slope of the tangent line to the graph of $y(x)$ in $(3,\ln(10))$ is $y'(3)=\tfrac{3}{5}$.
