I'm trying to use the Newton-Raphson in this function:
$$f(t)=0.5-e^{-\sqrt{2}t} \left(\cos\frac{2}{\sqrt2}t+\sin\frac{2}{\sqrt2}t \right) $$
I calculate the derivative with the product rule applying the product of the exponential with each trigonometric function, it yields
$$f(\dot{t})=\sqrt{2}e^{-\sqrt{2}t}\left(\cos \frac{2t}{\sqrt{2}}+ \sin \frac{2t}{\sqrt{2}}\right)+\frac{2e^{-\sqrt{2}t}}{\sqrt{2}}\left(\cos\frac{2}{\sqrt2}t-\sin\frac{2}{\sqrt2}t\right)$$
In the classroom it was said that the right answer is $$f\left(\dot{t}\right)=\frac{4}{\sqrt{2}}e^{-\sqrt{2}t}\sin\frac{2}{\sqrt{2}}t$$
but this was not explained why.
Which one is the right derivate? Or both are not good?
You have an error of sign:
$$\dot f=\sqrt{2}e^{-\sqrt{2}t}(\cos \frac{2t}{\sqrt{2}}+ \sin \frac{2t}{\sqrt{2}})\color{red}-\frac{2e^{-\sqrt{2}t}}{\sqrt{2}}(\cos\frac{2}{\sqrt2}t-\sin\frac{2}{\sqrt2}t)$$
than, with a bit of algebra, you can see that the result is the same as the classroom answer.