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Mathematics 13 Online
OpenStudy (anonymous):

what is the derivative of cosh^(3)t

OpenStudy (anonymous):

-3sinxcosx^2.... i think but i just learned the chain rule

OpenStudy (anonymous):

nope that's not it. I just checked :/ thank you for trying though!

OpenStudy (anonymous):

-3sinxcosx(^2)x

OpenStudy (anonymous):

its a hyperbolic function

OpenStudy (lgbasallote):

how about 3(cosht)^2(sinh)?

OpenStudy (lgbasallote):

3(cosh t)^2(sinh t) i meant

OpenStudy (anonymous):

I like that one too Igbasallote

OpenStudy (anonymous):

the diriveitive is cos is -sin... right?

OpenStudy (anonymous):

Not for hyperbolic cos functions. Just for regular cos functions. These have a little trick to them.

OpenStudy (anonymous):

[cosh^(3)t]' = [(cosh t)^3]' = 3*(cosh t)^2*[cosh t]' = 3*cosh^2(t)*sinh t

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

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