Derivative of Log Function: Is My Solution Correct?

[tbarnet7]

Homework Statement

y=ln (-2+√x)/(x^4)

Hi there. Just need some reassurance that I found the derivative of this equation correctly.

Also, if you can give me feedback on typing my equations that would also be wonderful. (Whether or not there are easier/better ways of typing the equations.)

Homework Equations

g(x)f'(x)-f(x)g'(x)/g(x)^2

The Attempt at a Solution

This is my final answer.
= x^(7/2) + 8x^3 +4x^3√x / 2x^8

 

[Mark44]

Homework Statement

y=ln (-2+√x)/(x^4)

Hi there. Just need some reassurance that I found the derivative of this equation correctly.

Also, if you can give me feedback on typing my equations that would also be wonderful. (Whether or not there are easier/better ways of typing the equations.)

Homework Equations

g(x)f'(x)-f(x)g'(x)/g(x)^2

The Attempt at a Solution

This is my final answer.
= x^(7/2) + 8x^3 +4x^3√x / 2x^8

I don't get this.

To make you work easier, rewrite the function like this:

$$y = \ln{\frac{-2 + x^{1/2}}{x^4}} = \ln(-2x^{-4} + x^{-7/2})$$
$$\frac{dy}{dx}= \frac{1}{-2x^{-4} + x^{-7/2}} \cdot \frac{d}{dx}(-2x^{-4} + x^{-7/2})$$

Can you continue from there?

I used LaTeX to format this stuff. To see what I did click either equation and a window will open with the LaTeX code.

 

[tbarnet7]

Thanks Mark. I checked out LaTeX and found it to be a bit further over my head at first glance so I will look into it deeper.

So; you multiplied the top by the recipricole of the bottom to get ln (-2x^-4 + x^ (7/2)) correct?

However; I am not sure how you got from their to the equation below. I am supposed to be applying the quotient rule. So maybe in answering your question; I don't know how to continue from there.

 

[Mark44]

Hold on a second - I might have misinterpreted what you wrote.

Is this your function?
$$\frac{\ln(-2 + \sqrt{x})}{x^4}$$
If so, my work doesn't reflect this.

Or is it this?
$$y = \ln{\frac{-2 + x^{1/2}}{x^4}}$$