Asked by Anonymous
Find dx/dy for the function y=x^cos(x)
Answers
Answered by
Steve
recall that if
y = a^v, y' = ln(a) a^v v'
y = u^n, y' = n u^(n-1) u'
So, interestingly enough, if
y = u^v, y' = vu^(v-1) u' + ln(u) u^v v'
so, here we have
y' = cosx x^(cosx-1) + ln(cosx) x^cosx (-sinx)
= x^cosx (cosx/x - ln(cosx) sinx)
y = a^v, y' = ln(a) a^v v'
y = u^n, y' = n u^(n-1) u'
So, interestingly enough, if
y = u^v, y' = vu^(v-1) u' + ln(u) u^v v'
so, here we have
y' = cosx x^(cosx-1) + ln(cosx) x^cosx (-sinx)
= x^cosx (cosx/x - ln(cosx) sinx)
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