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Frequently Asked Questions (FAQ)
What is the d/(dt)((e^{tb}-e^{ta})/(t(b-a))) ?
- The d/(dt)((e^{tb}-e^{ta})/(t(b-a))) is (be^{bt}t-ae^{at}t-e^{bt}+e^{at})/(t^2(b-a))
What is the first d/(dt)((e^{tb}-e^{ta})/(t(b-a))) ?
- The first d/(dt)((e^{tb}-e^{ta})/(t(b-a))) is (be^{bt}t-ae^{at}t-e^{bt}+e^{at})/(t^2(b-a))