I must find d/dx of the function in the picture. I get a different answer than the textbook. Can someone explain, based on my picture, what I did wrong?
The hyperbolic half-angle formula:
$\sinh(x) = \text{sgn}(x)\sqrt{\dfrac{\cosh(2x)-1}{2}}$
Square both sides, and you find that your answer is equal to the textbook's answer. Check out hyperbolic trig identities. Better yet, try to prove them on your own.