I read this in my textbook and I'm a bit confused by it:
So this is a bit confusing to me. A few questions:
When we have this line:
$$e^{(\ln b)x} \frac{d}{dx}(\ln b)x$$ $$= e^{(\ln b)x} \ln(b)$$
How does that happen? Don't we need to use product rule for $\frac{d}{dx}(\ln b)$$?
Why does $e^{(\ln b)x}$ = $b^x$
On
1)
$\frac {d}{dx}$ is the differential operator and it means differentiate whatever follows
$\frac {dy}{dx}$ means that we have applied the operation to the function $y(x)$
2)
We are applying the chain rule here.
If $y = f(g(x))$ then $\frac {dy}{dx} = f'(g(x))\cdot g'(x).$
And, in this case, $f(x) = e^x, g(x) = x\ln b$, and $f(g(x)) = e^{x\ln b}$
Maybe I am misunderstanding your question.
Product rule for $\frac {d}{dx} \ln b$ is not an issue bucause $\ln b$ is a constant and taking its derivative returns $0.$
3)
Why does $b^x = e^{x\ln b}$
Since $e^x$ and $\ln x$ are inverses of one another, $b = e^{ln b}$ and $b^x = e^{\ln(b^x)}$. And, $\ln b^x = x\ln b$ by the rules of logarithms.
You can use the product rule for $\dfrac d {dx}\big( (\ln b) x\big)$ just as you can for $\dfrac d {dx} (5x),$ thus: \begin{align} & \frac d {dx} (5x) = 5 \frac d {dx} x + x \frac d {dx} 5 = 5 \frac d{dx} x + (x\cdot 0) = 5\cdot 1 = 5. \\[12pt] & \dfrac d {dx}\big( (\ln b) x\big) = (\ln b) \frac d {dx} x + x \frac d {dx}(\ln b) = (\ln b) \frac d {dx} x + (x\cdot 0) = (\ln b) \cdot 1 = \ln b. \end{align} The derivative of a constant function is $0.$ Neither $b$ nor $\ln b$ changes as $x$ changes, so it is constant.
The difference between $\dfrac d {dx}$ and $\dfrac {dy}{dx}$ is that, for example, $$ \frac d {dx} \cos x = -\sin x \quad \text{ but }\quad \frac {dy}{dx} \cos x = \left( \frac d {dx} y\right) \cdot \Big( \cos x\Big). $$ In one case you're differentiating $\cos x$ with respect to $x,$ and in the other you're differentiating $y$ with respect to $x,$ where $y$ is some other function of $x.$