Suppose $u$ and $v$ are functions of $x$, and $f(x)=\ln (u v)+u^{2}$. If $u(1)=2, v(1)=7, u^{\prime}(1)=8$, and $v^{\prime}(1)=21$, then $f^{\prime}(1)=$

Step 1 ：Suppose $u$ and $v$ are functions of $x$, and $f(x)=\ln (u v)+u^{2}$. If $u(1)=2, v(1)=7, u^{\prime}(1)=8$, and $v^{\prime}(1)=21$, then we are asked to find $f^{\prime}(1)$.

Step 2 ：The derivative of $f(x)$ can be found using the chain rule and the product rule. The derivative of $\ln(uv)$ with respect to $x$ is $\frac{u'v+uv'}{uv}$ by the chain rule and the product rule. The derivative of $u^2$ with respect to $x$ is $2uu'$ by the chain rule. So the derivative of $f(x)$ is $\frac{u'v+uv'}{uv}+2uu'$.

Step 3 ：Substitute the given values of $u(1)$, $v(1)$, $u'(1)$, and $v'(1)$ into the derivative of $f(x)$ to find $f'(1)$.

Step 4 ：Final Answer: $f^{\prime}(1)=\boxed{39}$