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User Meriadec
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We want to find the derivative of


f(x)=11\sin x+4\cos x

To solve that we must remember the derivative of sin and cos


\begin{gathered} (d)/(dx)(\sin x)=\cos x \\ \\ (d)/(dx)(\cos x)=-\sin x \end{gathered}

Therefore, let's use it to find f'(x)


\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}

Therefore


f^(\prime)(x)=11\cos x-4\sin x

Now let's evaluate the derivative at the point 3π/4


\begin{gathered} f^(\prime)(3\pi/4)=11\cos(3\pi)/(4)-4\sin(3\pi)/(4) \\ \\ f^(\prime)(3\pi/4)=11(-√(2)/2)-4√(2)/2 \\ \\ f^(\prime)(3\pi/4)=-(11√(2))/(2)-(4√(2))/(2) \\ \\ f^(\prime)(3\pi/4)=-(1)/(2)(11√(2)+4√(2)) \\ \\ f^(\prime)(3\pi/4)=-(1)/(2)(15√(2)) \\ \\ f^(\prime)(3\pi/4)=-(15√(2))/(2) \end{gathered}

Hence


f^(\prime)(3\pi/4)=-(15√(2))/(2)

User Mrchad
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