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OpenStudy (anonymous):
what is the derivative of cosh^(3)t
14 years ago
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OpenStudy (anonymous):
-3sinxcosx^2.... i think but i just learned the chain rule
14 years ago
OpenStudy (anonymous):
nope that's not it. I just checked :/ thank you for trying though!
14 years ago
OpenStudy (anonymous):
-3sinxcosx(^2)x
14 years ago
OpenStudy (anonymous):
its a hyperbolic function
14 years ago
OpenStudy (lgbasallote):
how about 3(cosht)^2(sinh)?
14 years ago
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OpenStudy (lgbasallote):
3(cosh t)^2(sinh t) i meant
14 years ago
OpenStudy (anonymous):
I like that one too Igbasallote
14 years ago
OpenStudy (anonymous):
the diriveitive is cos is -sin... right?
14 years ago
OpenStudy (anonymous):
Not for hyperbolic cos functions. Just for regular cos functions. These have a little trick to them.
14 years ago
OpenStudy (anonymous):
[cosh^(3)t]' = [(cosh t)^3]' = 3*(cosh t)^2*[cosh t]' = 3*cosh^2(t)*sinh t
14 years ago
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OpenStudy (anonymous):
1/12(9sinh(x)+sinh3(x)) + C
because
formula says integral of cosh^m(x) = (sinh(x)cosh^(m-1) of (x))/m + ((m-1)/m)*integral of cosh^[m-2] of (x)
which becomes:
= 1/3 sinh(x) cosh^2(x)+2/3 integral cosh(x) dx
The integral of cosh(x) is sinh(x):
= (2 sinh(x))/3+1/3 sinh(x) cosh^2(x)+C
Which is equal to:
= 1/12 (9 sinh(x)+sinh(3 x))+C
14 years ago
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