Derivative Problems?

Question

Can anyone help me with these?

1. Given f(x)= log(base 3) (3-2x), f'(x)=_____

2. If y= ln abs(secx-tanx), then y'=______

3. If f(x)=(x^3)ln(x), then f'(x)=_______

4. derivative of ln e^2x=

5. If f(x)=ln(lnx), then f'(e) is...

6. If y=x^(x^3), then y'=_____

7. If y= e^(-x^2), then y"=______

8. If f(x)(1/3)(e^(3-2x)), then f'(x)=____

9. If x>0, then the derivative is (arcsin(1/x))=_____

PLEASE HELP ME!!!

i do not understand how to work the ln and logs to find derivatives...this course isnt being taught with a text book but through teacher notes and stufff...this is intro to tomorrows lesson

Answer 1

1. f(x) = log[base3](3 - 2x), use the chain rule.
f'(x) = (1/[(3-2x)ln3]) (-2)

2. y = ln |secx - tan x|
y' = 1/(secx - tanx) (sec(x)tan(x) - (sec(x))^2
y' = sec(x)

I'd help you with the rest but I firmly believe if you truly needed help, you wouldn't have put all of those questions to begin with. I highly advise you truly try learning instead of having people do homework for you.

Answer 2

1. 3^f(x) = 3^log(base 3)(3-2x) = 3-2x take ln both sides
ln(3)^f(x) = (3-2x) y*ln(3) = ln(3-2x) y = (1/ ln(3))*ln(3-2x) y' = 1/ln(3) *1/(3-2x)*(-2) = -2/ln(3) * 1/(3-2x)

Almost all of these can be solved using the product formula:

d(f*g)/dx = f*dg/dx + g*df/dx

or the chain rule

d(f(g(x)) = df/dg*dg/dx

or both.