# Homework Help: Derivatives with detailed pictures and work ! help !

1. Nov 20, 2008

### asdfsystema

deleted by accident .. ahh

Last edited: Nov 20, 2008
2. Nov 20, 2008

### Staff: Mentor

Speaking only for myself, I would be more inclined to help you on a single problem rather than a half dozen all at once.

3. Nov 20, 2008

### asdfsystema

good idea. i'll change it right now

4. Nov 20, 2008

### tiny-tim

Yup … that was what put me off a few hours ago!

ok, in 1. they want you to write it in the form log[f(x)] = …, and differentiate both sides.

And in 2., try using the chain rule … what do you get?

5. Nov 20, 2008

### asdfsystema

thank you tim,

Attempt at #1

log[f(x)]= log[ (x^2)(x-3)^9 / (x^2+3)^5 ]
log[f(x)] = log(x^2)(x-3)^9 - log (x^2+3)^5 ?

try#1
log[f(x)]= log(x^2)+log(x-3)^9-log(x^2+3)^5
i'm not sure if it is this ===== 2log(x) + 9log(x-3) - 5log(x^2+3)[/quote]
Yes, log(ab)= b log(a).
Now, differentiate.

and 2x/x^2= 2/x, of course: the derivative of 2log x is 2/x.

Are there 2 #2s?
No, the derivative of 5ln(u) is NOT 5 ln(u)*du/dx, it is (5/u)*du/dx.

Last edited by a moderator: Nov 20, 2008
6. Nov 20, 2008

### tiny-tim

hmm … messy, but mostly correct …

the lines I would pick out to string together for a proper answer are:

log[f(x)]= log[ (x^2)(x-3)^9 / (x^2+3)^5 ]

log[f(x)]= log(x^2)+log(x-3)^9-log(x^2+3)^5

= 2log(x) + 9log(x-3) - 5log(x^2+3)

then you should differentiate both sides (you only did the RHS):

d/dx(log[f(x)]) = 2x/x^2+ 9*1/x-3 - 5* 2x/x^2+3

and then complete that equation, and multiply both sides by … something … to get an equation with d/dx(f(x)) on the LHS on its own.
u= 5x+7ln(x)) du/dx= 5+ 7*1/x ==== 5+7/x is correct,

but the next line is completely wrong … try again!

7. Nov 20, 2008

### asdfsystema

hey guys,

1.

log[f(x)]= log[ (x^2)(x-3)^9 / (x^2+3)^5 ]

log[f(x)]= log(x^2)+log(x-3)^9-log(x^2+3)^5

= 2log(x) + 9log(x-3) - 5log(x^2+3)

d/dx(log[f(x)]) = 2x/x^2+ 9*1/x-3 - 5* 2x/x^2+3

1/f(x) * f'(x) = 2/x + 9/x-3 -10x/x^2+3

f'(x)= (2/x + 9/x-3 -10x/x^2+3) * (x^2)(x-3)^9/(x^2+5)^5

I got f(x) from the original equation

f'(3)= (2/3+9/0 - 30/12) * (9)(0)^9/(9+5)^5

Um ... is f'(3) = 9 ?

2.
I used d/dx 5ln(u) = 5/u * du/dx

so continuing off from there, 5/5x+7ln(x) * 5+7/x is the answer?

f'(3)= (5/5(3))+7ln(3) * 5+7/3

==== 1/3 + 7.6902 * 7.3333
final answer f'(3) = 56.7287 ?

Thanks , I tried to make it as clear as possible this time haha :)

8. Nov 20, 2008

### tiny-tim

Much better!

(though you have a misprint at the end … 5 instead of 3 )

Well, you have 09/0 …

wouldn't it have been better to cancel an (x - 3) before you put x = 3?
Yes … much clearer!

But clarity isn't just for the reader's benefit … it's for your benefit also …

so your next aim should be to use brackets more often … in fact, whenever they're needed … you've made at least one mistake because you've misread a non-existent bracket.

9. Nov 20, 2008

### asdfsystema

thank you , i will do that more often !

1. sorry i made a mistake last time , i meant f'(3) = 0 (the 9 is too close to the 0)

(2/x + 9 - 10x/x^2+3) * ( (x^2)((x-3)^8)/ (x^2+3)^5 =

(2/3 + 9 - 30/12) * ( (9)*(0) / (12)^5 ) = 0 because 9*0 = 0

Last edited: Nov 20, 2008