Question

if y= e^x-e^-x then Dy/dx-√y^2+4 is equal to​

Answer 1

ZERO is the answer.

$y = {e}^{x} - {e}^{ - x} \\ \frac{dy}{dx} = {e}^{x} + {e}^{ - x} \\ \frac{dy}{dx} - \sqrt{( {y}^{2} - 4)} \\ {e}^{x} - {e}^{ - x} - \sqrt{( { {e}^{x} - {e}^{ - x} )}^{2} + 4} \\ {e}^{x} - {e}^{ - x} - \sqrt{ { ( { {e}^{x} })^{2} + ({e}^{ - x}) }^{2} - 2 + 4 } \\ {e}^{x} - {e}^{ - x} - \sqrt{ { {(e}^{x} + {e}^{ - x} ) }^{2} } \\ 0$

Hope this helps you.

Answer 2

Step-by-step explanation:

hope the attached pic help uhh....