Masaken
Unknown Member
Use mathematical induction to prove that the nth derivative of y = cos x is y^n = cos(x+{[n*pi]/2}). (sorry, not able to ss the qn itself but the expression is in the attached ss of the working out)
Basically I got this question wrong and got stuck when I was attempting to redo it. I've attached the working out for the question, however I don't get it; in the first one, why is n = 1 true? Is it simply because the derivative of cos is -sin?
![1663993261256.png](/data/attachments/36/36343-0e2d53bcc28e7ebb51f2955460c8f03d.jpg)
![1663993272550.png](/data/attachments/36/36344-c062be94ac02bee16a2cb205ddae9072.jpg)
And in the second ss above, when I attempted the inductive step for n = k+1, while I did y=k^(n+1), I broke it apart and tried to do y^k * y^1; why did they do [y^k]^1? Wouldn't that just be y^k?? Thanks in advance
Basically I got this question wrong and got stuck when I was attempting to redo it. I've attached the working out for the question, however I don't get it; in the first one, why is n = 1 true? Is it simply because the derivative of cos is -sin?
![1663993261256.png](/data/attachments/36/36343-0e2d53bcc28e7ebb51f2955460c8f03d.jpg)
![1663993272550.png](/data/attachments/36/36344-c062be94ac02bee16a2cb205ddae9072.jpg)
And in the second ss above, when I attempted the inductive step for n = k+1, while I did y=k^(n+1), I broke it apart and tried to do y^k * y^1; why did they do [y^k]^1? Wouldn't that just be y^k?? Thanks in advance