Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Find the derivative of xe^-2x

Mathematics
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

@rajathsbhat can u help?
can you do it if i teach you how to derive e^-2x?
I got 2e^-2x is that correct?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

product rule..
okay
what you got when you differentiated the exponent is correct......
okay
do you know the product rule??
no
this is the product rule..... it says if you have eg.... \[y=uv\] \[\frac{ dy }{ dx }= uv \prime + vu \prime \]
ok
can you give an example?
yes...sure eg. \[y= 7xe ^{x ^{2}}\] \[\frac{ dy }{ dx }=7x \times 2xe ^{x ^{2}} +xe ^{x ^{2}} \times 7\]
ok
can you simplify it further from there?????
i think so
thx for the help
you are welcome.....

Not the answer you are looking for?

Search for more explanations.

Ask your own question