If \(f(1)=1, f^{\prime}(1)=5\), then the derivative of \(\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^2\) at \(x=1\) is

\(\text { Let } y =\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^2 \)
\( \therefore \frac{\mathrm{~d} y}{\mathrm{~d} x} =\mathrm{f}^{\prime}(\mathrm{f}(\mathrm{f}(x))) \cdot \mathrm{f}^{\prime}(\mathrm{f}(x)) \cdot \mathrm{f}^{\prime}(x)+2 \mathrm{f}(x) \cdot \mathrm{f}^{\prime}(x) \)
\( \therefore \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{x=1} =\mathrm{f}^{\prime}(\mathrm{f}(\mathrm{f}(1))) \cdot \mathrm{f}^{\prime}(\mathrm{f}(1)) \cdot \mathrm{f}^{\prime}(1) \)
\( =\mathrm{f}^{\prime}(\mathrm{f}(1)) \cdot \mathrm{f}^{\prime}(1) \cdot 5+2 \cdot 1 \cdot 5 \)
\( =\mathrm{f}^{\prime}(1) \cdot 5 \cdot 5+2 \cdot 1 \cdot 5 \)
\( =5 \cdot 5 \cdot 5+2 \cdot 1 \cdot 5 \)
\( =135\)