How am I doing on derivatives??

[Jade]

I want to make sure I am getting the hang of scary looking deravitives. I do not need to simlify.

1) d/dx{(3e^(2x^2-x) + 2)^30}

My answer: 30(3e^(2x^2-x) + 2)^29 (3e^(2x^2-x))(4x-1)

2) d/dx ((e^-3x)/(lnx))

My answer: [(e^-3x)(-3)(lnx) - (e^-3x)(1/x)]/(lnx)^2

3)d/dx (ln(sqrt 'x' + 3))

My answer: This one seems a little more difficult. the square root x is confusing me. I am thinking you should change it into ln sqrt 'x' + ln3 before you take derivative and then it would be whatever the derivative of ln sqrt 'x' is + 1/3?

 

[galactus]

3)d/dx (ln(sqrt 'x' + 3))

My answer: This one seems a little more difficult. the square root x is confusing me. I am thinking you should change it into \(\displaystyle \L\\\underbrace{ln sqrt x + ln3}_{\text{no, you can't.}}\) before you take derivative and then it would be whatever the derivative of ln sqrt 'x' is + 1/3?

Above, you can't do that. You're mixing up your log rules. If it were \(\displaystyle ln(3\sqrt{x})=ln(\sqrt{x})+ln(3)\), you could.


Actually, this one isn't bad. Just chain rule and remember what the derivative of ln(x) is.....1/x.

\(\displaystyle \L\\ln(\sqrt{x}+3)\)

\(\displaystyle \frac{d}{dx}[ln(u)]du=(\frac{1}{u})u'\)

\(\displaystyle \L\\(\frac{1}{\sqrt{x}+3})(\frac{1}{2\sqrt{x}})\)

 

[Jade]

That wasn't so bad

Thanks