Im trying to use the Newton-Rhapson in this function:

f(t)=0.5−e−√2t(cos2√2t+sin2√2t)f(t)=0.5-e^{-\sqrt{2}t}(cos\frac{2}{\sqrt2}t+sin\frac{2}{\sqrt2}t)

I calculate the derivative with the product rule applying the product of the exponential with each trigonometric function, it yields

f(˙t)=√2e−√2t(cos2t√2+sin2t√2)+2e−√2t√2(cos2√2t−sin2√2t)f(\dot{t})=\sqrt{2}e^{-\sqrt{2}t}(cos \frac{2t}{\sqrt{2}}+ sin \frac{2t}{\sqrt{2}})+\frac{2e^{-\sqrt{2}t}}{\sqrt{2}}(cos\frac{2}{\sqrt2}t-sin\frac{2}{\sqrt2}t)

But in the classroom it was said that the right answer is

f(˙t)=4√2e−√2tsin2√2tf(\dot{t})=\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?

Thanks in advance

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1 Answer

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You have an error of sign:

˙f=√2e−√2t(cos2t√2+sin2t√2)−2e−√2t√2(cos2√2t−sin2√2t)\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.