(eu)'=ueu-1
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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