Obnoxious Posted September 15, 2005 Posted September 15, 2005 [math]f(x) = e^x[/math] I keep getting the silly answer of just X, when I know it is [math]e^x[/math]
mezarashi Posted September 15, 2005 Posted September 15, 2005 The question is what again? Given a function, you want to derive that function. Or do you mean differentiation using first principles? What have you tried so far. Show what you've done to get your answer.
Obnoxious Posted September 15, 2005 Author Posted September 15, 2005 I was given [math]y = e^x[/math] So I did this! [math]\ln y = x \ln e[/math] [math]dx/dy = 1/y[/math] [math]dy/dx = y[/math] Am I making some illegal moves?
Klaynos Posted September 15, 2005 Posted September 15, 2005 I think I understand what you are trying to do, show that f(x) = ex And f'(x) = ex Are both true? [math]\frac {\ln y} {\ln e} = x[/math] You need to do that stage before you can differentiate it wrt y
DQW Posted September 15, 2005 Posted September 15, 2005 I was given[math]y = e^x[/math] So I did this! [math]\ln y = x \ln e[/math] [math]dx/dy = 1/y[/math] [math]dy/dx = y[/math] Am I making some illegal moves? If you are allowed to use the derivative of the logarithm (which is no more fundamental than that of the exponent), you have correctly found the answer, dy/dx = y = e^x If you have to prove this from "more fundamental" results, I suggeswt you use the series expansion for e^x and differentiate term by term.
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