# Thread: strange step used in solutions

1. ## strange step used in solutions

i wont write the whole question, but the term
$\frac{dy}{dx} \times \frac{1}{y}$
in the next line they change to
$\frac{d (ln(y))}{dx}$

i just dont see why you would be able to do that. none of the other terms in the equation (i havent written out) are changed. only thing i can think of is some sort of implicit differentiation idea or some sort. im not too sure

2. Originally Posted by walleye
i wont write the whole question, but the term
$\frac{dy}{dx} \times \frac{1}{y}$
in the next line they change to
$\frac{d (ln(y))}{dx}$

i just dont see why you would be able to do that. none of the other terms in the equation (i havent written out) are changed. only thing i can think of is some sort of implicit differentiation idea or some sort. im not too sure
Without the whole question there is no context in which to judge why this is done. However, if you're question is that you don't understand the equality of the two expressions, then you should note that from the chain rule it follows that

$\frac{d (\ln y)}{dx} = \frac{d (\ln y)}{dy} \cdot \frac{dy}{dx}$ ....

3. Originally Posted by walleye
i wont write the whole question, but the term
$\frac{dy}{dx} \times \frac{1}{y}$
in the next line they change to
$\frac{d (ln(y))}{dx}$

i just dont see why you would be able to do that. none of the other terms in the equation (i havent written out) are changed. only thing i can think of is some sort of implicit differentiation idea or some sort. im not too sure
It is the chain rule of differentiation.

If I asked you to differentiate $\ln(y)$ with respect to $x$, you'd have to do it like so:

$\frac{d}{dx} \ln(y) = \frac{dy}{dx} \times \frac{d}{dy} \ln(y) = \frac{dy}{dx} \frac{1}{y}$

4. ahhh i see, thanks to both of you

Originally Posted by mr fantastic
Without the whole question there is no context in which to judge why this is done.
yeah its obvious (to me) why they do it, that step just confused me is all

thanks