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OpenStudy (laddiusmaximus):
find the derivative of y=xsec kx i thought since it was asking for dy/dx, that k would be zero since its a constant, but i was wrong.
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OpenStudy (laddiusmaximus):
my answer was y' = (secx tanx)(sec kx) but I was a bit off.
OpenStudy (anonymous):
yes it is incorrect
it should be
y'=sec(kx)+x*(sec(kx)(tan (kx))*k
OpenStudy (anonymous):
product rule
OpenStudy (laddiusmaximus):
well the book has sec kx(kx tan kx +1) i thought chain rule would be used, but the k is confusing me plus I dont know where the plus one came from
zepdrix (zepdrix):
look at psi's answer a sec, :)
If you factor a sec(kx) out of each term, you get what the book has :D
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OpenStudy (anonymous):
dont rely on books, they save space as it costs a lot for printing long expressions :)
OpenStudy (anonymous):
be self confident
OpenStudy (laddiusmaximus):
so product rule? I dont need to use the chain rule?
OpenStudy (anonymous):
both
OpenStudy (laddiusmaximus):
so product rule first, then chain rule?
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OpenStudy (anonymous):
yes depends , how you use it though
OpenStudy (anonymous):
did you solved it yet? what answer you got?
OpenStudy (anonymous):
dy/dx=kx tan kx sec kx + sec kx
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