When I calculate the derivative of $\displaystyle y = sinh(x) tanh(x) $ I come up with $\displaystyle y' = cosh(x) tanh(x) + sinh(x) sech^2(x) $ Should this be further simplified?
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$\displaystyle cosh(x) \cdot tanh(x)=sinh(x)$ then take $\displaystyle sinh(x)$ as a common factor ..
So... $\displaystyle y' = sinh(x)(1+sech^2(x)) $ Sorry, it has been a very long time since I have had to work these sorts of problems.
Correct .
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