Time Derivative of ln: Solving $\frac{d}{dx} ln x$

[lavster]

Homework Statement

what is the time derivative of the following expression: $$ln(\frac{ct+\sqrt{(c^2t^2-s^2)}}{s})$$

Homework Equations

$$\frac{d}{dx} ln x = \frac{1}{x}$$

chain rule

log rules eg ln (x/y) = ln x - ln y

The Attempt at a Solution

$$ln(ct+\sqrt{(c^2t^2 -s^2)})-ln s$$

so time derivative is:

$$<br /> \frac{1}{ct+\sqrt{(c^2t^2-s^2)}}(\frac{1}{2}(c^2t^2-s^2)^{-\frac{1}{2}}2c^2t+c)-0$$

using the above formulae. however the answer says $$\frac{1}{c\sqrt{c^2t^2-s^2}}...$$


can anyone help?

thanks

 

[Mark44]

Homework Statement

what is the time derivative of the following expression: $$ln(\frac{ct+\sqrt{(c^2t^2-s^2)}}{s})$$

Homework Equations

$$\frac{d}{dx} ln x = \frac{1}{x}$$

chain rule

log rules eg ln (x/y) = ln x - ln y

The Attempt at a Solution

$$ln(ct+\sqrt{(c^2t^2 -s^2)})-ln s$$

so time derivative is:

$$<br /> \frac{1}{ct+\sqrt{(c^2t^2-s^2)}}(\frac{1}{2}(c^2t^2-s^2)^{-\frac{1}{2}}2c^2t+c)-0$$

using the above formulae. however the answer says $$\frac{1}{c\sqrt{c^2t^2-s^2}}...$$

I don't see anything wrong with what you did. I suspect that all that you need to do is some algebraic simplification.