Question

I need help question

Answer 1

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)`