probability - Is it true in general that $E(1/X) = 1/E(X

Differentiate the equation of the curve. y = (e^x)/x. y' = (e^x)/x - (e^x)/x^2. To solve for the slope, substitute 1 in place of x, since is the x-coordinate of the given point.[/has_googlemeta5][has_googlemeta6]. Move slider below to add more terms... 3 . Find the Inverse Function f(x)=(1+e^x)/(1-e^x) Replace with . Interchange the variables. Solve for . Tap for more steps... Rewrite the equation as . Solve for . Tap for more steps... Multiply each term by and simplify. Tap for more steps... Multiply each term in by . Cancel the common factor of .

How to integrate 1/1+e^x - Quora

Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.. A lot of good answers here, but I have another approach: [math]{\displaystyle \int{\frac{1}{1+e^x}}dx}[/math] We will rewrite our numerator as: [math]1+e^x-e^x[/math] This rewriting of the numerator is algebraically legal as the adding and subtrac... . Series Expansion. A series expansion is a representation of a particular function as a sum of powers in one of its variables, or by a sum of powers of another (usually elementary) function .

Is e^-x = 1/e^x? - Quora

Yes, e^-x = 1/e^x Proof: Let P denote the product of e^-x and e^x Then P = e^-x . e^x = e^ (-x+x) = e^0,……………………………(1) since according to the. I'm finding the critical points of f(x) = e x - x. First step: Find f'(x) f'(x) = e x - 1 Second Step: Set to 0 and solve for x e x = 1 My question now is how to solve for x.. Find the differential dy when y=e^(x/10) And evaluate dy and the change of y if x=0 and dx=0.1 I have no idea where to begin, if someone could list the steps to this problem, and general steps to solving problems like this it would be much appreciated. Thank you for any help.