Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can you tell me if there are any flaws with this derivation of $y = a^x$...

The assumptions are that the derivative $$\frac{d}{dx}e^x = e^x$$ and that the derivative $$\frac{d}{dx}\ln x = \frac{1}{x} = x^-1.$$

$$\begin{array}{rcl} y &=& a^x\\ \ln y &=& \ln a^x\\ \ln y &=& x \ln a\\ x &=& \frac 1{\ln (a) }\ln y\\ \frac {dx}{dy} &=& \frac {1}{\ln (a) } y^-1\\ \frac {dy}{dx} &=& \ln(a)y\\ \frac {d}{dx} a^x &=& \ln(a)a^x \end{array} $$ I find the natural logarithm in this result very odd, a bit random.

share|improve this question
    
Looks all good to me. –  user157545 Jun 25 at 6:30
1  
If $a=e$, then we have the familiar $y=e^x$ and its derivative $$\frac{d}{dx} e^x=\ln(e) e^x=1\cdot e^x=e^x.$$ –  le gâteau au fromage Jun 25 at 6:34
1  
Looks good, except... differentiation –  Mar Johnson Jun 25 at 7:10
    
A nitpick, use curly brackets like x^{-1} instead of x^-1 as you used... Otherwise very nice! –  Joachim Jun 25 at 12:58

1 Answer 1

up vote 6 down vote accepted

Your work is correct. Notice also that we can differentiate this function by simpler way:

$$y=a^x=e^{\ln(a) x}$$ so using the chain rule $$\frac{dy}{dx}=\ln(a)e^{\ln(a) x}=\ln(a) a^x$$

share|improve this answer
    
Bravo! That is clever. Thanks. –  Colin Jun 25 at 6:46
    
You're welcome. –  Sami Ben Romdhane Jun 25 at 6:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.