Hey guys the question asks to prove d/dt (Sin^-1(e^t)) = (e^-2t-1)^-1/2 I've got that the derivative is e^T/(1-e^2t)^1/2 but how do I get that to be the right hand side of the equation?
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Originally Posted by Mathsnewbie Hey guys the question asks to prove d/dt (Sin^-1(e^t)) = (e^-2t-1)^-1/2 I've got that the derivative is e^T/(1-e^2t)^1/2 but how do I get that to be the right hand side of the equation? Multiply your last expression by $\displaystyle \frac{\displaystyle\frac{1}{\sqrt{e^{2t}}}}{\displ aystyle\frac{1}{\sqrt{e^{2t}}}}$
Can you explain to me how that works out? I multiply both the top and bottom by that but still don't get any furhter
Originally Posted by Mathsnewbie Can you explain to me how that works out? I multiply both the top and bottom by that but still don't get any furhter Use the chain rule on the left hand side to differentiate and get $\displaystyle \frac{e^t}{\sqrt{1 - e^{2t}}}$. Now use Chop'S hint to show that the the right hand side is equal to this expression.
I think I must be multiplying it out wrong, I get when I multiply it out ((E^2t)^1/2).E^t/ (E^2t-(E^2t)^2)^1/2 From there I can't figure out a way to simplify further.
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