calculus 2

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CalcGuy

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working on chp 6 transcendental functions specificaly 6.8 the inverse trig functions and their derivatives...

(im not sure how to get this in any other way)

the integral evaluated from 0 to 3 or is it from 3 to 0? the 3 is on top and the 0 is on the bottom... of e^2x times cos(e^2x) dx

i think i got it... i think its .0617

u=e^2x
du=2e^2x dx

then i get 1/2 (integral symbol) cos(u) du

which is 1/2 sin(u) + C

plug in u...

right?
 
\(\displaystyle \int_0^3 {e^x \cos \left( {e^x } \right)dx} = \mathop {\left. { \sin \left( {e^x } \right)} \right|}\nolimits_0^3\)
 
right how did u type that in?

however mine is e^2x so i would end up with 1/2 sin(e^2x)|3 and 0 right?
 
ok i have another one... y=5ln(6arccosh x)... find y'...

so i...

let u=(6arc cosh x) and du= 6/the square root of ((x^2) -1)

so then...

y=5ln(u)

y'=5/u du

plug in u and du

and i get an ugly ans which makes me think its wrong

i get...

5/[(arc cosh x)(the square root of ((x^2) -1))]
 
Yep. That looks good. See, you're getting the hang of it.

BTW, that is LaTex we are using to make things display nicely. To see the code used click on 'quote' in the upper right hand corner of the post.

\(\displaystyle y'=\frac{5}{cosh^{-1}\sqrt{x^{2}-1}}\)
 
hmm that quote looks harder than calc

let me see if i can modify it

i got the same thing u typed but i have an extra x on the denominator for arc cosh x .... is my ans wrong?

\(\displaystyle y'=\frac{5}{xcosh^{-1}\sqrt{x^{2}-1}}\)[/quote]
 
awww... i found a similar problem in the book... u 4got ur x... my ans was right
 
The derivative of \(\displaystyle 5ln(6cosh^{-1}(x))=\frac{5}{\sqrt{x^{2}-1}cosh^{-1}(x)}\)

I stand by it. There is no reason why there should be an extra x in the denominator.

An extra x creates a whole nuther non-integrable function.
 
\(\displaystyle f(x) \ = \ 5ln[6cosh^{-1}(x)] \ = \ 5[ln(6)+ln[cosh^{-1}(x)]] \ = \ 5ln(6)+5ln[cosh^{-1}(x)]\)

\(\displaystyle f' (x) \ = \ 0 \ + \ 5 \ \bigg(\frac{1/\sqrt(x^{2}-1)}{cosh^{-1}(x)}\bigg) \ = \ \frac{5}{\sqrt(x^{2}-1)cosh^{-1}(x)}\)

\(\displaystyle Note: \ \frac{d[cosh^{-1}(u)]}{dx} \ = \ \frac{u'}{\sqrt(u^{2}-1)}\)
 
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