d (e^(r ln x) )/ dx = [e^(r ln x)] [d (r ln x) / dx] = x^r (r / x) = r x^(r−1)?
Could someone explain to me how is it that [e^(r ln x)] [d (r ln x) / dx] = x^r (r / x) = r x^(r−1)
its the proof in my calculus 2 book for how even irrational numbers follow the derivative rules of nX^(n-1)
I dont get how they got rid of e^(r ln x) I can see that it equals e^r but they don't have that,
Update:I also don't see how they went from x^r (r/x) to r x^(r-1) they must be using properties of logs but I cant figure it out.
More Questions From This User See All
Answers & Comments
for clarity in bookkeeping, let us label the function y = r ln x, so that what you have is
d / dx (e^y)
= (dy/dx) e^y , chain rule
now
dy/dx = d/dx (r ln x) = r d/dx (ln x) = r / x
and
e^y = e^{r ln x} = e^{ ln (x^r) } = x^r, so
d / dx (e^y) = x^r (r / x) = r (x^r / x) = r x^(r - 1) which is what you have.
The final step was writing x^r / x = x^r x^{-1} = x^{r + (-1) } = x^{r - 1}, there are no logs here, this is a property of powers.
Division: x^a / x^b = x^(a - b)
Multiplication: x^a x^b = x^(a + b)
Power to a power: (x^a)^b = x^(ab)
The ln and e functions are inverses of each other, one "undoes" the other (when applied together they equal 1 by the definition of an inverse), more generally if we have some function w, and inverse w^(-1) is defined such that w w^(-1) = w^(-1) w = 1. For natural logarithms and exponentials, we have
ln (e^x) = x ln (e) = x*1 = x
and
e^(ln x) = x <-- to show this one takes a little more formality which might not be so useful to see, but let us know if you want to see it.
So, e^(r ln x) does not equal e^r, but x^r, since you can rewrite: r ln x = ln (x^r), apply an exponential (the logarithms inverse): e^(ln (x^r)) = x^r.
Does this make sense?
Ummm what