coth(x) = cosh(x)/sinh(x) = (ex + e-x)/ (ex - e-x) or (e2x + 1)/ (e2x - 1)
I assume you mean e2x. I tried this at the Wolframalpha site (using the search term: "integrate e^(2*x)"; the site says it is e2x/2. This can be obtained by a simple substitution.
If you mean ex squared, the answer is e2x
It os 1/2*e2x + c
Derivative of e2x - y = 2e2x - dy/dx
Original function: f(x) = e2x Integrated function: F(x) = e2x/2 Evidence: F'(x) --> f(x) = e2x ; F'(x) = 2e2x/2 = e2x = f(x) Q.E.D
3
e2x=ex^2 basically means that x2=2x, in which case x2-2x=0, x = 0, 2. I don't think that's what the question meant. It could mean: e2x=(ex)2 . Which comes from one of the rules of exponents. Basically, look at it this way: Take the natural log of both sides: ln e2x= ln(ex)2 From rules of logs: (2x) ln e = (2) ln ex 2x ln e = (2) (x) ln e 2x = (2)(x)
A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)A polynomial equation: ax4+ bx3+ cx2+ dx + e = 0A trigonometric equation: sin(3x+2) = 0 Combinations: cos(x3+ e2x) = ln(x)
y = e2x+1
This question involves a property of dividing common bases. For example, xm/xn is the same as saying x(m-n). Whenever two common bases are being divided, the exponents can be combined by subtraction. So, for this problem: e(2x-1)/e(x-1), the e can be written once, with the exponents being subtracted: e2x-1-(x-1), distributing the minus sign arrives at e2x-1-x+1, or ex. ANS: ex
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