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

calvin140404

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Jomo

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

polskabritva

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\(\displaystyle \left(45-45\,e^{-\frac{\ln\left(90\right)\,x}{2}}\right)'_x=\dfrac{45\,\ln\left(90\right)\,e^{-\frac{\ln\left(90\right)\,x}{2}}}{2}\)
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HallsofIvy

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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 \(\displaystyle -45e^{-\frac{ln(90)x}{2}}\)

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 \(\displaystyle e^{-\frac{ln(90)x}{2}}\).

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

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

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