#leibniz rule using H'(t) substitution

1 messages · Page 1 of 1 (latest)

drowsy zenith
#

I'm having trouble with the method I've been taught from here involving using function H to remove the need to do multiple integrals, in the original equation there is an integral at the end which isn't used in the method and is needed to gain the correct answer

image.png
gray martenBOT
#
  1. Do not ping the Moderators, unless someone is breaking the rules.
  2. Do not ping the Helper Moderators, unless there is a conflict between helpers.
  3. Do not ping other members randomly for help.
  4. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
  5. Wait patiently for a helper to come along.
  6. If the Helper has answered your question, remember to thank them with the Mathematics Ranks bot and close the thread with:

+close
Feel free to nominate the person for helper of the week in #helper-nominations
If you're happy with the help you got here, and the server overall, you can contribute financially as well:

drowsy zenith
#

this being the question

image.png
#

the answer is 0. however using the method I get $$\frac{sin(1)-sin(2)}{x}$$

dim ospreyBOT
drowsy zenith
#

using the actual equation it seems you get this:

image.png
#

I have the first 2 parts of the 1st line of the equation however I'm missing the last part

#

My Working is:
$$\frac{d}{dx} \int_{1/x}^{2/x} H'(t)dt = \frac{d}{dx} (H(t)|^{2/x}_{1/x}) = \frac{d}{dx}(H(2/x)-H(1/x))$$
then differentiate:
$$H'(2x^{-1})(-\frac{2}{x^{2}})-H'(x^{-1})(-\frac{1}{x^{2}})$$
Then I get:
$$\frac{H'(x^{-1})-2H'(2x^{-1})}{x^{2}}$$

dim ospreyBOT
oak oak
#

Well, use the substitution xt=u

elfin pagodaBOT
#

@drowsy zenith

<:HelpIcon:1304095958283321385>| Help Reminder

Hello yooda., this is a friendly reminder that your help request has been inactive for more than 24 hours. If you no longer need assistance, please consider closing the thread using the +close command. This thread will be automatically closed in 3 days if it remains inactive.

drowsy zenith
#

+close

elfin pagodaBOT
# drowsy zenith +close
Do you still want to close your help request?

No eligible helpers were found in this thread. You can still close this post if you don't require helper any longer.