Again, I'm trying to follow an example in my textbook.
Basically, we're trying to find the derivative with respect to $x$ of
$$y(x)=[A(x)]^\alpha[B(x)]^\beta[C(x)]^\gamma$$
According to the book, the first step is to take the natural logarithm, so:
$$ln(y)=\alpha*ln(A(x))+\beta*ln(B(x))+\gamma(ln(C(x))$$
I'm onboard so far, but in the next step, the book states that the derivative with respect to x is:
$$\frac{y'}{y}=\alpha\frac{A'(x)}{A(x)}+\beta\frac{B'(x)}{B(x)}+\gamma\frac{C'(x)}{C(x)}$$
The explanation goes on for a while longer, but this step leaves med dumbfounded. How and why is this possible?
Hint:
That step is using the chain rule.
For instance, if you have $y=\log\left(f(x)\right)$, then differentiating with respect to $x$, we have:
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}\left(\log\left(f(x)\right)\right)}{\mathrm{d}x} \cdot \frac{\mathrm{d}\left(f(x)\right)}{\mathrm{d}x}$$
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{f(x)} \cdot \frac{\mathrm{d}\left(f(x)\right)}{\mathrm{d}x}$$
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{f'(x)}{f(x)}$$
In general if you have $F(x)=f(g(x))$, then differentiating with respect to $x$ you will get:
$$F'(x)=f'(g(x))g'(x)$$