Asked by KKK
                 how does thederivative of -2xcos(x^2)= -4xcos(x^2) -2sin(x^2) and not -4x^2cos(x^2) -2sin(x^2)?
            
            
        Answers
                    Answered by
            oobleck
            
    Huh? If you let
u = x^2, du = 2x dx, so
∫ -2x cos(x^2) dx = ∫ -cosu du = -sinu = -sin(x^2)
I suspect a typo.
    
u = x^2, du = 2x dx, so
∫ -2x cos(x^2) dx = ∫ -cosu du = -sinu = -sin(x^2)
I suspect a typo.
                    Answered by
            Reiny
            
    if y = (-2x)(cos(x^2))
then the derivative dy/dx = (-2x)(-2x)(sin(x^2)) + (-2)cos(x^2) , by the product rule
= (4x^2)sin(x^2) - 2cos(x^2)
Neither of your choices matches that, so like oobleck, I suspect a typo.
    
then the derivative dy/dx = (-2x)(-2x)(sin(x^2)) + (-2)cos(x^2) , by the product rule
= (4x^2)sin(x^2) - 2cos(x^2)
Neither of your choices matches that, so like oobleck, I suspect a typo.
                    Answered by
            oobleck
            
    oops. I thought it said anti-derivative. Mental typo!
    
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.