Answer
$\frac{\lambda_0 + 6\lambda^5}{2-\lambda_0}$
Work Step by Step
$\lambda_0$ is a constant by assumption, so the entire denominator is a constant.
\begin{align}
\frac{d}{d\lambda} \bigg[ \frac{\lambda \lambda_0 + \lambda^6}{2 - \lambda_0}\bigg] \\
& = \frac{d}{d\lambda}\bigg[\frac{\lambda_0}{2-\lambda_0} \lambda + \frac{1}{2-\lambda_0} \lambda^6\bigg] \\
&= \frac{\lambda_0}{2-\lambda_0} + \frac{6}{2-\lambda_0} \lambda^5 \\
&= \frac{\lambda_0 + 6\lambda^5}{2-\lambda_0} \\
\end{align}
