$f(x)=a^x=(a)^x=(e^{\ln a})^x=e^{x \ln a}$
$f'(x)=$?
I don't quite understand how the Chain Rule is applied here and what is the inner\outer function? Could someone elaborate?
PS Oops, I meant natural exponential function.
$f(x)=a^x=(a)^x=(e^{\ln a})^x=e^{x \ln a}$
$f'(x)=$?
I don't quite understand how the Chain Rule is applied here and what is the inner\outer function? Could someone elaborate?
PS Oops, I meant natural exponential function.
I like dy/dx notation and substitution whenever the chain rule comes up.
$y = a^x = e^{xln(a)}.$
$u = xln(a) \implies \dfrac{du}{dx} = ln(a).$
$And\ u = x * ln(a) \implies y = e^u \implies \dfrac{dy}{du} = e^u = e^{xln(a)} = a^x \implies \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} \implies$
$\dfrac{dy}{dx} = a^x * ln(a).$
You mention inner and outer functions. I assume you mean the chain rule (with appropriate assumptions) for the derivative of $f(g(x))$ is $f^{\prime}(g(x))g^{\prime}(x)$. Remember an alternate notation for $e^x$ is $\exp(x)$. So $a^x=e^{x\ln(a)}=\exp(x\ln(a))$. Let $f(x)=\exp(x)$ and $g(x)=x\ln(a)$. So the derivative is $f^{\prime}(g(x))g^{\prime}(x)=\exp(x\ln(a))\cdot \ln(a)=a^x\ln(a)$
Or, rather e depends on it's convoluted exponent same as $e^x$ depends on x, when we differentiate... so we just rewrite e with its convoluted exponent, much the same as we would do if we defferentiated $e^x$ over x?
$w=x^2+3$
$\dfrac{d}{dw} e^{x^2+3}=e^{x^2+3}$
because
$\dfrac{d}{dx} e^x=e^x$
I have no idea what a convoluted exponent is. Perhaps it is what happens to an exponent that has been out drinking with contortionists.
But yes. (Note: I truly wish that, in beginning explanations. math texts would not use the same letter to refer to different variables.)
$y = e^x \implies \dfrac{dy}{dx} = e^x.$ That is a basic derivative, a general rule, just like the addition, product, and power rules.
$\therefore w = u^2 + 3\ and\ z = e^{(u^2 + 3)} \implies z = e^w\ and \dfrac{dz}{dw} = e^w = e^{(u^2 + 3)}.$ We applied the general rule about powers of e.
$But\ w = u^2 + 3 \implies \dfrac{dw}{du} = 2u.$ We applied the general rules about the derivatives of sums and the derivatives of a variable to a power.
$\therefore \dfrac{dz}{du} = \dfrac{dz}{dw} * \dfrac{dw}{du} = e^{(u^2 + 3)} * 2u = 2ue^{(u^2 + 3)}.$ We applied the chain rule.
The chain rule:
$(f \circ g)'(u) = f'(g(u))\cdot g'(u)$
For the function:
$h(x) = a^x = (e^{\log(a)})^x = e^{(\log(a))x}$,
let $f(x) = e^x$ and let $g(x) = (\log(a))x$ <--$g(x)$ is just a constant (in this case, the constant is $\log(a)$) times $x$, that is $g(x) = cx$, where $c = \log(a)$.
Then $h = f \circ g$.
Applying the chain rule, we have $f'(x) = e^{x}$, and $g'(x) = c = \log(a)$.
So at the point $x = u$, we have $g(u) = (\log(a))u$, and $f'(g(u)) = e^{g(u)} = e^{(\log(a))u}$, and we have $g'(u) = c = \log(a)$.
To get the derivative of $h = f \circ g$ at the point $u$, we multiply these two numbers together:
$h'(u) = (e^{(\log(a))u})\cdot \log(a)$.
We can simplify this a bit, since (reversing the steps we took earlier), $e^{(\log(a))u} = (e^{\log(a)})^u = a^u$, so that $h'(u) = a^u \cdot \log(a)$.
So the derivative of $h$, as a function of $x$, is given by:
$h'(x) = a^x\cdot\log(a)$.
Note that gives us the same answer as before, when $a = e$:
Writing $e^x$ as $\exp(x)$, we have:
$\exp'(x) = e^x\cdot \log(e) = e^x\cdot 1 = e^x = \exp(x)$.