finding derivative

[crazyvirus85]

i can't find the derivative of

f(x) = ln[sqrt((x-9)/(x-7))]

i got -8/(x^2+16x+63) but it's not right......... :evil:

 

[stapel]

I'm sorry, but we can't help you find your error until you show your steps.

Eliz.

 

[crazyvirus85]

f'(x) = (1/(sqrt((x-9)/(x-7))) * (1/2) (((x-9)/(x-7))^(-1/2)) * ((1)(x-7)-(x-9)(1))/(x-7)^2

(1/2) * 1/((x-9)/(x-7)) * -16/(x-7)^2

(1/2) * ((x-7)/(x-9)) * -16/(x-7)^2

-16/(2(x-9)(x-7))

-8/(x^2 - 16x + 63)

 

[stapel]

I agree with your first line:

dy/dx = [1/sqrt([x - 9]/[x - 7])] (1/2)[1/sqrt([x - 9]/[x - 7])] [1(x - 7) - 1(x - 9)]/[x - 7]²

But I don't follow you from then on. My first step would be to simplify that last numerator and combine the two square roots:

. . . . .dy/dx = (1/2)(1/[(x - 9)/(x - 7)]) [(x - 7 - x + 9)/(x - 7)²]

. . . . .dy/dx = [(x - 7)/(2(x - 9))] [2/(x - 7)²]

The 2's cancel out to give:

. . . . .dy/dx = [(x - 7)/(x - 9)] [1/(x - 7)²]

Cancel off the duplicate factor.


Eliz.

 

[tkhunny]

I think we may have crossed over into insanity.

f(x) = ln[sqrt((x-9)/(x-7))] = ½[ln(x-9) - ln(x-7)]

Isn't that about a billion times easier?

 

[stapel]

Isn't that about a billion times easier?

Oh... are we supposed to, like, use those log rule thingies we learned back in algebra?

Yes, you're quite right; that would have been a much more intelligent first step. Thank you!

Eliz.