We want to find the derivative of
[tex]f(x)=11\sin x+4\cos x[/tex]To solve that we must remember the derivative of sin and cos
[tex]\begin{gathered} \frac{d}{dx}(\sin x)=\cos x \\ \\ \frac{d}{dx}(\cos x)=-\sin x \end{gathered}[/tex]Therefore, let's use it to find f'(x)
[tex]\begin{gathered} f^{\prime}(x)=[11\sin(x)+4\cos(x)\rbrack^{\prime} \\ \\ f^{\prime}(x)=[11\sin(x)\rbrack^{\prime}+\lbrack4\cos(x)\rbrack^{\prime} \\ \\ f^{\prime}(x)=11\cdot[\sin(x)\rbrack^{\prime}+4\cdot\lbrack\cos(x)\rbrack^{\prime} \\ \\ f^{\prime}(x)=11\cdot\cos x-4\sin x \end{gathered}[/tex]Therefore
[tex]f^{\prime}(x)=11\cos x-4\sin x[/tex]Now let's evaluate the derivative at the point 3π/4
[tex]\begin{gathered} f^{\prime}(3\pi/4)=11\cos\frac{3\pi}{4}-4\sin\frac{3\pi}{4} \\ \\ f^{\prime}(3\pi/4)=11(-\sqrt{2}/2)-4\sqrt{2}/2 \\ \\ f^{\prime}(3\pi/4)=-\frac{11\sqrt{2}}{2}-\frac{4\sqrt{2}}{2} \\ \\ f^{\prime}(3\pi/4)=-\frac{1}{2}(11\sqrt{2}+4\sqrt{2}) \\ \\ f^{\prime}(3\pi/4)=-\frac{1}{2}(15\sqrt{2}) \\ \\ f^{\prime}(3\pi/4)=-\frac{15\sqrt{2}}{2} \end{gathered}[/tex]Hence
[tex]f^{\prime}(3\pi/4)=-\frac{15\sqrt{2}}{2}[/tex]