Jon P. answered • 02/26/15
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Do you mean t sin t / (1 + t)? That's what it looks like, so I'll assume that's it.
From the way you've written the last expression, you're not keeping track of parentheses.
The derivative is ((1 + t) Dx t sint - (t sin t)(1)) / (1 + t)2
= ((1 + t) (t cos t + sin t) - t sin t) / (1 + t)2
= (t cos t + sin t + t2 cos t + t sin t - t sin t) / (1 + t)2
= (t cos t + sin t + t2 cos t) / (1 + t2)
And that's as far as the simplification will go.
What does the book say?
Jon P.
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Well let's see what their numerator comes out to:
(t2+1)(cost+sint) = t2 cos t + t2 sin t + cos t + sin t
Our numerator is t cos t + sin t + t2 cos t.
Let's subtract ours from theirs. If we can make it come out to 0, then the two expressions are equivalent.
t2 cos t + t2 sin t + cos t + sin t - (t cos t + sin t + t2 cos t) =
t2 sin t + cos t - t cos t
I may be missing something, but I don't see any way to make that 0. So I am as confused as you are.
Sometimes the book is wrong.
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02/26/15
Dreazie M.
I have been working this same problem (that's why I looked it up; my answer didn't match the one in the back of Stewart Calculus either) and I think I have figured out how to make our numerator look like the book's
start with ours which is written:
t^2 cos t + t cos t + sin t
factor out cos t to leave:
cos t(t^2 + t) + sin t
then write that in a different order so that it says:
(t^2 +t)cos t + sin t
and of course we put that over the denominator of (1 + t)^2 and it becomes
(t^2 +t)cos t + sin t / (1 + t)^2
which is what the answer says in the book :) THAT'S F*N TEAM WOOORK!
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03/02/16
Hannah G.
02/26/15