derivative of a logarithmic function

[Elisabeth V kessel]

Hello dear everyone,
I am trying to determine the derivative of the following logaritmic function, the answer is given but I do not understand how to come to this specific solution. Therefore I really hope you could help me out. I do not understand how they come to 3.ln (x^2+4.x)/2. It would be amazing if someone knew how te arrive at that point and what the steps inbetween are.

 

[yoscar04]

Take note that you can combine x²+4x with the second term having the square root.

 

[Steven G]

[MATH]\dfrac{d}{dx}\ln[f(x)] = \dfrac{f'(x)}{f(x)}[/MATH]

 

[Steven G]

BTW, I get f'(2) = 3/2

 

[JeffM]

There is less to go wrong if you "see" the u-substitution.

[MATH]y = ln\{(x^2 + 4x) * \sqrt{x^2 + 4x}\} \text { and } u = x^2 + 4x.[/MATH]
[MATH]\therefore y = ln\{u * u^{1/2}\} = ln(u^{3/2}) = \dfrac{3}{2} ln(u) \text { and } \dfrac{du}{dx} = 2x + 4.[/MATH]
[MATH]\therefore \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = \dfrac{3}{2} * \dfrac{1}{u} * (2x + 4) =[/MATH]
[MATH]\dfrac{3(2x + 4)}{2(x^2 + 4x)}.[/MATH]
There is NO NEED to do the substitution. I just find it cuts down on transcription errors (like the one your book made). The more things you need to copy, the greater the chance of error.