HWilliams44
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Another integral/derivative problem
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pka
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None of what you did in correct.
What is the derivative of \(\displaystyle \arcsin \left( {{e^x}} \right)~?\)
Thank you for showing your work.
Let \(\displaystyle u = e^x,\;\;\;\;\;\;\dfrac{du}{dx} = e^x,\;\;\;\;\;\;dx = \dfrac{du}{u}\)
After substituting, the integral becomes
\(\displaystyle \displaystyle \int \dfrac{3u}{\sqrt{1 - u^2}}\dfrac{du}{u} = \int \dfrac{3}{\sqrt{1 - u^2}}\ du\)
That is as far as that u-substitution will take you. If you don't recognize the integral, check for trig function inverses.
There is no justification for removing the square root. Lets look closely at the u-substitution.
Let \(\displaystyle u = e^x,\;\;\;\;\;\;\dfrac{du}{dx} = e^x,\;\;\;\;\;\;dx = \dfrac{du}{u}\)
After substituting, the integral becomes
\(\displaystyle \displaystyle \int \dfrac{3u}{\sqrt{1 - u^2}}\dfrac{du}{u} = \int \dfrac{3}{\sqrt{1 - u^2}}\ du\)
That is as far as that u-substitution will take you. If you don't recognize the integral, check for trig function inverses.
HWilliams44
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Now I get it! So, you would get 3arcsin(e^x), right? Thanks for all the help, you guys! 