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derivative of a^x with respect to x

tmt

Active member
Jan 15, 2014
236
I thought that when y = a^x,

dy/dx= a^x

But I just read somewhere that it is in fact a^x lnx.

Which one is it?

Thanks!
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
I thought that when y = a^x,

dy/dx= a^x

But I just read somewhere that it is in fact a^x lnx.

Which one is it?

Thanks!
d/dx (e$^x) = e^x$

as a^x = e$^{ln a x}$ you can proceed
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
If:

$f(x) = a^x$ (for this to make sense, we require $a > 0$)

we can write:

$f(x) = (g \circ h)(x)$

where:

$h(x) = \ln(a)x$ <---this is just multiplication by a constant
$g(x) = e^x$

since:

$(g\circ h)(x) = g(h(x)) = g(\ln(a)x) = e^{\ln(a)x} = (e^{\ln(a)})^x = a^x$.

The CHAIN RULE then states:

$f'(x) = g'(h(x))h'(x)$.

Now:

$g'(x) = e^x$
$h'(x) = \ln(a)$

So:

$f'(x) = (e^{\ln(a)x})(\ln(a)) = a^x\ln(a)$

rather than the $a^x\ln(x)$ you posted.

In other words, when we change the "base" of an exponential function, we have to modify the derivative by a constant factor, and this constant factor is the (natural) logarithm of the base.

When considering "exponential growth", what we mean is that the rate of growth is directly proportional to the population:

$f(x) = kf'(x)$.

These functions are of the form:

$f(x) = Ae^{kx}$

If we write $k = \ln(b)$ for some positive number $b$ (we can always do this, since the range of $\ln$ is the entire real numbers), we see that:

$f(x) = Ae^{\ln(b)x} = Ab^x$

in other words, "changing the exponent" of an exponential function by a linear factor (it being a linear factor is important, here), is really the same thing as "changing the base".

So...it really doesn't make too much sense to investigate "different bases", since this is pretty much the same thing as investigating how $e^{kx}$ behaves, for different $k$.

A good "self-test" for seeing how well you understand all this, is to try to differentiate:

$f(x) = x^x$

(you will need to use the chain rule AND the product rule).