Need some help deriving this e^x function. Its quite a complex one.

No, this is actually an easy problem. Don't you realize that ln(0.545)2-\frac{\ln(\dfrac{0.5}{45})}{-2} is just a constant?

How would you find the derivative of y= 3e^(2x) and y = -4e^(11x)? How about y = a*e^(kx) where k is some constant?
 
First, the derivative of f+ g is the derivative of f plus the derivative of g and the derivtive of a constant, 45, is 0 so this is the same as the derivative of just 45eln(90)x2\displaystyle -45e^{-\frac{ln(90)x}{2}}

Second, the derivative of a constant, -45, times f is that constant times the derivative of f so that is -45 times the derivative of eln(90)x2\displaystyle e^{-\frac{ln(90)x}{2}}.

Third the derivative of e to the f power is e to the f power times the derivative of f so we have 45eln(90x)2\displaystyle -45e^{-\frac{ln(90x)}{2}} times the derivative of ln(90)x2=(ln(90)2)x\displaystyle \frac{ln(90)x}{2}= \left(\frac{ln(90)}{2}\right)x.

And that, again, is a constant times x so its derivative is just ln(90)2\displaystyle -\frac{ln(90)}{2}.

So the derivative of 45eln(90)x2+45\displaystyle -45e^{-\frac{ln(90)x}{2}}+ 45 is
(45)(ln(90)2)eln(90)x2\displaystyle (-45)\left(-\frac{ln(90)}{2}\right)e^{-\frac{ln(90)x}{2}}
 
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