TuringTest
  • TuringTest
Let\[y=x^p(1+x)^q\]find\[y^{(p+q)}\]
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

TuringTest
  • TuringTest
that's the (p+q)th derivative by the way Sorry that I have to go soon, but I look forward to seeing your responses when I return.
amistre64
  • amistre64
ln(y) = ln(x^p) + ln(1+x)^q y'/y = p/x + q/(1+x) hmm
anonymous
  • anonymous
i think the answer is just (p+q)!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
since the highest power of x will be x^(p+q), and all the other powers of x will become 0 after P+q derivatives.
amistre64
  • amistre64
sounds plausible to me
TuringTest
  • TuringTest
hint:(Leibniz' theorem/ binomial stuff)
TuringTest
  • TuringTest
at least in MIT's solution I'd like to see others
amistre64
  • amistre64
lol, id like to see mits solution :)
TuringTest
  • TuringTest
I'll send it to you....
AccessDenied
  • AccessDenied
im curious, what class is this problem from?
TuringTest
  • TuringTest
OCW calc I
AccessDenied
  • AccessDenied
Ahh, okay. Thank you. :)
Mr.Math
  • Mr.Math
http://en.wikipedia.org/wiki/Leibniz_rule_(generalized_product_rule)

Looking for something else?

Not the answer you are looking for? Search for more explanations.