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

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.