#$&*
Mth 271
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Derivative of exponential
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At the conclusion of my previous question you stated that the derivative of y(x)=A*b^x is y`(x)=A*b^x *ln(b).
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I understand that this is an exponential function. The typical way i have so far seen an exponential function expressed is y=e^x the derivative of which is e^x. Because the inverse of y=e^x is y=ln(x), we can prove the derivative of an exponential function to be that function by proving that the derivative of ln(x) = 1/x. I have seen this proof and understand it fairly well, but I am unsure as to how this is applied in proving the derivative of an exponential function.
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More specifically, how does your statement above, derivative of y(x)=A*b^x is y`(x)=A*b^x *ln(b) relate to the function y=e^x and how do we prove this?
How is the derivative of ln(x) used to prove that the derivative of e^x is e^x. Is there another way to prove this?
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Excellent questions.
It's easier to prove directly that the derivative of e^x is e^x.
The proof of this is fairly challenging but not too much so. The derivative is the limitng value of
(e^(x + `dx) - e^x) / `dx
as `dx -> 0.
The numerator can be written
e^x ( e^`dx - 1).
For small values of `dx, we can show that e^(`dx) is very close to 1 + `dx, so that the difference quotient for small enough value of `dx is very close to
e^x * `dx / `dx,
which is equal to e^x.
We need to show that e^(`dx) is approximately equal to 1 + `dx.
This is proven using the definition of e, which is
e = limit{n -> infinity}( (1 + 1/n)^n),
along with the formula for the binomial expansion.
To prove that the derivative of ln(x) is 1/ x we use the chain rule along with the fact that the derivative of e^x is e^x.
The formula for the derivative of b^x relies on the fact that b = e^(x ln(b)), which follows directly from the fact that the natural log is inverse to e^x. (From this fact we conclude that e^(x ln(b)) = (e^ln(b))^x = b^x).
Combining this with the chain rule we reach our conclusion.
We will encounter the chain rule, and learn more about manipulating the (1 + 1/n)^n expression, a little later in the course. At that time this should all make very good sense to you.
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#$&*
Mth 271
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
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Once we've proved that the derivative of y=A*b^x is y`=A*b^x *ln(b), how do we use this function to find the average rate between two clock times of a function of the form P(t) = P0 * (1 + r) ^ t?
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Once i had the derivative I would substitute the clock time values, say 2 and 4, into the rate function respectively and then average these two values by adding them and dividing by two. I believe there is another way to do this.
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Once we have the derivative of the exponential function, how do we use this function to find the average rate of change between two clock times?
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The average rate of change is just
ave rate = (change in P) / (change in t) = (P(t_2) - P(t_1)) / (t_2 - t_1).
Given the values of, or expressions for, t_1 and t_2 you just substitute and simplify.
By the rule for the derivative of the exponential function b^x, using (1 + r) for b, we find that the derivative of P(t) is
P ' (t) = P0 * (1 + r)^t * ln(1 + r).
Given an interval from t_1 to t_2, you could evaluate P(t) at t_1 and t_2, then average the values to get an approximate average rate of change. Multiplying this by the duration of the interval, which is (t_2 - t_1), you would get the approximate change in the value of the function.
This is another great question, and your insights are excellent.
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