Asked by Anonymous
find f'(x)
(sinx)^cosx
(sinx)^cosx
Answers
Answered by
bobpursley
See the generalized power rule here:
http://en.wikipedia.org/wiki/Table_of_derivatives
http://en.wikipedia.org/wiki/Table_of_derivatives
Answered by
drwls
Use the chain rule for differentiating a function of a function.
Let cosx = v(x)
d/dx[(sinx)^v(x)]= d(sinx)^v/dv * dv/dx
= d/dv[sqrt(1-v^2)]/dv * dv/dx
= (-2)(1/2)/[sqrt(1-v^2)] * (-sin x)
= (-1)/sin x * (-sin x) = 1
A very interesting result!
Let cosx = v(x)
d/dx[(sinx)^v(x)]= d(sinx)^v/dv * dv/dx
= d/dv[sqrt(1-v^2)]/dv * dv/dx
= (-2)(1/2)/[sqrt(1-v^2)] * (-sin x)
= (-1)/sin x * (-sin x) = 1
A very interesting result!
Answered by
drwls
My answer looks rather suspicious, although I verified it at x = 0 and x = 0.5. I hope that other teachers will respond.
Answered by
Reiny
I would do it this way:
take ln of both sides
lny = ln(cosx^sinx)
= sinx(ln(cosx))
(dy/dx)/y = sinx(cosx/sinx) + (-sinx)(ln(sinx)
dy/dx = (cosx)^sinx[sinx(cosx/sinx) + (-sinx)(ln(sinx)]
don't know how much simplication is needed.
take ln of both sides
lny = ln(cosx^sinx)
= sinx(ln(cosx))
(dy/dx)/y = sinx(cosx/sinx) + (-sinx)(ln(sinx)
dy/dx = (cosx)^sinx[sinx(cosx/sinx) + (-sinx)(ln(sinx)]
don't know how much simplication is needed.
Answered by
Reiny
Got my sines and cosines mixed up as I copied from paper.
try again:
lny = ln(sinx^cosx)
= cosx(ln(sinx))
(dy/dx)/y = cosx(cosx/sinx) + (-sinx)(ln(sinx))
dy/dx = dy/dx = (sinx)^cosx[cosx(cosx/sinx) + (-sinx)(ln(sinx)]
try again:
lny = ln(sinx^cosx)
= cosx(ln(sinx))
(dy/dx)/y = cosx(cosx/sinx) + (-sinx)(ln(sinx))
dy/dx = dy/dx = (sinx)^cosx[cosx(cosx/sinx) + (-sinx)(ln(sinx)]
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.