Q: What is the derivative of e to the u?

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the derivative of tangent dy/dx [ tan(u) ]= [sec^(2)u]u' this means that the derivative of tangent of u is secant squared u times the derivative of u.

the derivative of Ln(u)=u'/u

y"+y'=0 is a differential equation and mean the first derivative plus the second derivative =0.Look at e-x the first derivative is -e-xThe second derivative will be e-xThe sum will be 0

If dy/dx = (e) (9x) then Y = 4.5ex2 plus (any constant).==================================The above answer explains how to get the integral of e9x.If you were interested in how to get the derivative of e9x, the answer is e9.I suspect you may have actually wanted to ask how to get the derivative of e9x.In that case, the derivative of e9x is 9e9x.

d/dx e3x = 3e3x

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The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)

The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)

the derivative of tangent dy/dx [ tan(u) ]= [sec^(2)u]u' this means that the derivative of tangent of u is secant squared u times the derivative of u.

ee is a constant and so its derivative is 0.

e^(-2x) * -2 The derivative of e^F(x) is e^F(x) times the derivative of F(x)

2.71828183 ==So the derivative of a constant is zero.If you have e^x, the derivative is e^x.

The derivative of xe is e. The derivative of xe is exe-1.

e is not derivative. Please reconsider your question, clarify what you wish to ask and resubmit.

The derivative of ex is ex

The first derivative of e to the x power is e to the power of x.

You must first apply the product rule to the term: 23te Product Rule: y=uv= u(dv/dx) + v(du/dx) *the derivative of e^x is still e^x and since your 'x' value is one, the derivative of 'e ^1' stays as 'e' plug it in: y = 23t (e) + 23 (e) then include the derivation of your last term: -1.2t ---> -1.2 answer: y = 23te + 23e -1.2

the derivative of Ln(u)=u'/u