# Thread: proof of derivative of a^x

1. ## proof of derivative of a^x

how do you prove that the derivative of a^x = (a^x)(ln a)?

i tried using logarithmic differentiation:
y = a^x
ln y = ln (a^x)
ln y = x (ln a)
d/dx (ln y) = d/dx [x (ln a)]
(1/y) (dy/dx) = (1)(ln a) + (x)(1/a)
dy/dx = y [(ln a) + (x/a)]
so if i substitute
dy/dx = a^x [(ln a) + (x/a)]

did i do something wrong? how come it's not coming out correctly? i used logarithmic differentiation to prove that the derivative of e^x = e^x and it worked. how come it doesn't work for proving this?

2. Originally Posted by oblixps
y = a^x
ln y = ln (a^x)
ln y = x (ln a)
d/dx (ln y) = d/dx [x (ln a)]
(1/y) (dy/dx) = (1)(ln a) + (x)(1/a)
dy/dx = y [(ln a) + (x/a)]
so if i substitute
dy/dx = a^x [(ln a) + (x/a)]
?
You're differentiating with respect to $\displaystyle x$. So, $\displaystyle \ln a$ is just simply a constant which you can pull out, i.e.: $\displaystyle \frac{d}{dx} \left(x \ln a\right) = \left(\ln a\right) \left[ \frac{d}{dx} \big( x \big)\right]$

3. oh wow i can't believe i didn't see that...
thanks!!

4. i tried using logarithmic differentiation:
y = a^x
ln y = ln (a^x)
ln y = x (ln a)
d/dx (ln y) = d/dx [x (ln a)]
(1/y) (dy/dx) = (1)(ln a) + (x)(1/a)
dy/dx = y [(ln a) + (x/a)]
so if i substitute
dy/dx = a^x [(ln a) + (x/a)]

did i do something wrong? how come it's not coming out correctly?
YES you did.
BUT a is a constant, so there's no product rule here
$\displaystyle {d\over dx}(x\ln a)=\ln a$
Now continue...

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# proof Dx of a^x

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