# Finding Derivatives with Respect to a Variable

1. Mar 27, 2012

### ialan731

1. The problem statement, all variables and given/known data

The question says ti find the derivative of y with respect to the independent variable.
The equation is: y=4 ln 2t.

2. Relevant equations

I know how to find derivatives using the product/quotient/chain/etc rules, but that isn't what the question is asking for. I though it wanted me to use implicit differentiation, but that doesn't really make sense because y is explicitally defined in terms of x. I don't really know what should be used.

3. The attempt at a solution

Before, I tried to just get the derivative normally and got 4 ln 2t +2/t, but that's not right. Then I also tried using implicit differentiation, but it wasn't working.

By the way, the answer is (ln4/t)4 ln 2t

Thank you in advance!

2. Mar 28, 2012

### QuarkCharmer

What's the derivative of ln(x) ?

If you know the derivative of that, then you can simply apply the chain rule.

$$\frac{dy}{dx} = 4\frac{d}{dx}ln(2x)$$

recall that $$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$$

3. Mar 28, 2012

### ialan731

The derivative of ln(x) is 1/x.

Am I supposed to use implicit differentiation, then? Or is the chain rule enough?

4. Mar 28, 2012

### QuarkCharmer

The chain rule is plenty.

What you can do is simply use a substitution. Let some function U(t) equal the "contents" of the original (ln) function.

Then find the derivative of ln(U), then apply the chainrule.

5. Mar 28, 2012

### ialan731

The derivative of ln(U) would just be 1/U then. I would only be working with the equation 4ln(U). After I use the chain rule for that equation, would I substitute the original ln function?

6. Mar 28, 2012

### QuarkCharmer

4 is a constant so it just comes outside, then let f(x) = ln(U), and let U(x) = 2t then simply recall the chain rule:

$$4\frac{df}{dx} = 4\frac{df}{du} \frac{du}{dx}$$ //I just multiplied in a 4 there for the constant

7. Mar 28, 2012

### ialan731

Okay. I get it now. This would work for all problems like this, right?

8. Mar 28, 2012

### QuarkCharmer

Well yes. Technically you ALWAYS are using the chain rule. It's just that for cases like the derivative of y=x (y'=1), the "inner product" or composite function's derivative is just 1 also.

If you have a ton of functions, let's say something like this:

y = x, x = t, t = p

You can see that you could just substitute it all in, or you can simply follow the chain rule:

$$\frac{dy}{dp} = \frac{dy}{dx} \frac{dx}{dt} \frac{dt}{dp}$$

and so on, up to an infinite number of nestled composite functions.

9. Mar 28, 2012

### ialan731

Great! This was the one thing I was worried about, but it doesn't seem to be too bad now. Thank you!

10. Mar 28, 2012

### QuarkCharmer

What solution did you get?

11. Mar 28, 2012

### ialan731

I got (ln4/t)4 ln 2t.