• Nov 8th 2006, 05:27 PM
nobodygirl
Hi, I really hate to ask for help as I'm usually good at math, but I have homework due at 10 o'clock tonight and I'm struggling. Could someone please help me with the following problems? I'm not asking for you to do my homework for me as that wouldn't help me get a good grade in the class, I just have no idea how to do these.

3. The next one is REALLY confusing.

F(x)=f(x^3) and G(x)=(f(x))^3

a^2=13
f(a)=3
f'(a)=11
f'(a^3)=15

I found F'(a), but I can't figure out G'(a)

I really hate to ask so many questions, but I've been working on this for an hour and I've gotten nowhere on these. Help on ANY of them would be great.

Thank you.
• Nov 8th 2006, 06:07 PM
AfterShock
Quote:

Originally Posted by nobodygirl
Hi, I really hate to ask for help as I'm usually good at math, but I have homework due at 10 o'clock tonight and I'm struggling. Could someone please help me with the following problems? I'm not asking for you to do my homework for me as that wouldn't help me get a good grade in the class, I just have no idea how to do these.

1. lnx^5 and I need to find the first and second derivatives

Let f(x) = [ln(x)]^5; I am assuming you mean this? Or, do you mean ln(x^5)?

I'll do both; I'll assume for case 1 you mean the first and for case 2 you mean the latter:

Case 1:

f(x) = [ln(x)]^5
f'(x) = 5*[ln(x)]^4*(1/x) = [5*[ln(x)]^4]/x
f''(x) = Use quotient rule:

Quotient rule states:

[f(x)/g(x)]' = [g(x)*f'(x) - f(x)*g'(x)]/[g(x)]^2

[x*20*[ln(x)]^3*(1/x) - [5*[ln(x)]^4]*1]/[x]^2

= [20*[ln(x)]^3 - 5*[ln(x)]^4]/x^2

Case 2:

f(x) = ln(x^5)
f'(x) = 5/x
f''(x) = -5/(x^2)

I think the above is pretty easy to see, so I think you mean the first one.
• Nov 8th 2006, 06:14 PM
AfterShock
Quote:

Originally Posted by nobodygirl
Hi, I really hate to ask for help as I'm usually good at math, but I have homework due at 10 o'clock tonight and I'm struggling. Could someone please help me with the following problems? I'm not asking for you to do my homework for me as that wouldn't help me get a good grade in the class, I just have no idea how to do these.

5. find f'(x) when f(x)=-6cos(sin(x^6))

This is actually easier than it looks; we have a composition of functions.

f(x) = -6cos(sin(x^6))

f'(x) = -6*-sin(sin(x^6))*cos(x^6)*6*x^5

Simplify:

Thus, f'(x) = 36*x^5*cos(x^6)*sin(sin(x^6))
• Nov 8th 2006, 06:16 PM
nobodygirl
I was talking about the first way.

I tried putting in the answer you had for that one, and it said that it was incorrect. My question is how did you get the 5ln^4 times 1/x?
• Nov 8th 2006, 06:17 PM
nobodygirl
Quote:

Originally Posted by AfterShock
This is actually easier than it looks; we have a composition of functions.

f(x) = -6cos(sin(x^6))

f'(x) = -6*-sin(sin(x^6))*cos(x^6)*6*x^5

Simplify:

Thus, f'(x) = 36*x^5*cos(x^6)*sin(sin(x^6))

Ah, thanks, I just made several stupid mistakes in that one.
• Nov 8th 2006, 06:19 PM
AfterShock
Quote:

Originally Posted by nobodygirl
Hi, I really hate to ask for help as I'm usually good at math, but I have homework due at 10 o'clock tonight and I'm struggling. Could someone please help me with the following problems? I'm not asking for you to do my homework for me as that wouldn't help me get a good grade in the class, I just have no idea how to do these.

4. I need to find the f'(x) when f(x)=8e^(xcosx)

f(x) = 8e^(xcosx)

Another easy one;

f'(x) = 8*e^(x*cos(x))*{derivative of inside of composition here = g(x)}

g(x) = x*cos(x)
g'(x) = 1*cos(x) - x*sin(x)

Thus, f'(x) = 8*e^(x*cos(x))*(cos(x) - x*sin(x))
• Nov 8th 2006, 06:20 PM
nobodygirl
Quote:

Originally Posted by AfterShock
f(x) = 8e^(xcosx)

Another easy one;

f'(x) = 8*e^(x*cos(x))*{derivative of inside of composition here = g(x)}

g(x) = x*cos(x)
g'(x) = 1*cos(x) - x*sin(x)

Thus, f'(x) = 8*e^(x*cos(x))*(cos(x) - x*sin(x))

I ALWAYS forget product rule, thank you so much, you're a life saver!
• Nov 8th 2006, 06:23 PM
AfterShock
Quote:

Originally Posted by nobodygirl
I was talking about the first way.

I tried putting in the answer you had for that one, and it said that it was incorrect. My question is how did you get the 5ln^4 times 1/x?

If you integrate both of them, you will see they are correct. That is, you will end up with f'(x) integrating f''(x) and f(x) integrating f'(x).

f(x) = [ln(x)]^5
f'(x) = 5*[ln(x)]^4*(1/x) = [5*[ln(x)]^4]/x

(1/x) is taking the derivative of the inside function; do you see why?

5*[ln(x)]^4*{derivative of composition here}, in which case we need to find the derivative of ln(x), which is 1/x, thus,

f'(x) = 5*[ln(x)]^4*(1/x) = [5*[ln(x)]^4]/x
• Nov 8th 2006, 06:30 PM
nobodygirl
Quote:

Originally Posted by AfterShock
If you integrate both of them, you will see they are correct. That is, you will end up with f'(x) integrating f''(x) and f(x) integrating f'(x).