Jtdylktuy

Show that
$$
frac{d}{dx} a^x = a^x ln a.
$$

How would I do a proof for this. I can't seem to get it to work anyway I try.

I know that
$$
frac{d}{dx} e^x = e^x.
$$
Does that help me here?

Hint: $a^x=e^{ln a^x}=e^{xln a}$.

Let $f : mathbb{R} to mathbb{R}$ be given by $f(x)=a^x$ and consider the $ln$ function. We can take the composition so that we have:

$$(lncirc f)(x)=ln (a^x)=xln a$$

Now, if we take the derivative, on the left hand side we use the chain rule and on the right hand side we differentiate as usual so that we have:

$$frac{f'(x)}{f(x)}=ln a$$

Now solving for $f'(x)$ gives $f'(x) = f(x) ln a$ so that $f'(x) = a^x ln a$. This useful technique can be used to take derivatives of other functions: we compose the original function with the inverse and then differentiate on both sides and use the same idea we've used here, this technique can simplify many derivatives and save a lot of time in some situations.

Hint: Write $y = a^x$ or equivalently $ln y = x ln a$ and use implicit differentiation.

, the last term, , can be explained by L'Hopital's rule, just taking the limits of the numerator and denominator one of each time.

Here are the steps
$$
frac{d}{dx} left[a^xright] = frac{d}{dx} left[e^{ln a^x}right]= e^{ln a^x} frac{d}{dx} left[ln a^xright]
$$
$$
= a^x frac{d}{dx} left[xln aright] = a^xleft(ln aright)frac{d}{dx} left[xright]= a^xln a
$$

$ frac{d}{dx} a^x=$

$=d/dx$ $e^{ln (a^x)}=$

$=d/dx e^{xln(a)}$ just as you said.

Now it's time for chain rule.

$f(x)=e^x$

$g(x)=x*ln(a)$

$a^x= f(g(x))$

last comment before I get solving

$a=e^{ln(a)}$

now let's get it done

$d/dx e^{x ln(a)}=$

$=e^{x*ln(a)} d/dx (x ln(a))=$ (by chain rule)

$=e^{x ln(a)} * ln(a)$

and that is the solution. Wait it doesnt look correct --> we should do some algebra!

Show for yourself that

$e^{x*ln(a)}=( e^{ln(a)} )^x$ but as we said that $a = e^{ln(a)}$ so $( e^{ln(a)} )^x$ actually equals $a^x$ and of course $e^{x*ln(a)}$ equals a^x

With what we just said you can see that $e^{x * ln(a)} ln(a)$ equals $a^x ln(a)$ or

$$d/dx a^x = a^x ln(a)$$

Let $y = a^x$

Then taking log on both side to the base $e$

We have;

$ ln y = ln a^x$

$ ln y = xcdot ln a$

Taking derivative with respect to $x$;

$frac d{dx} ln y = frac d{dx} xcdot ln a$

(I have applied chain rule here>>)

$ frac 1y frac{dy}{dx} = ln a + 0$

$ frac {dy}{dx} = ycdot ln a$

We know $y= a^x$

So, $frac {dy}{dx} = a^x ln a$

Hope it was easier than other methodologies used to derive it!