# Differentiation Help

1. Jun 21, 2006

### Hollysmoke

Greetings. I was just wondering if someone could take a look and tell me I did the following correctly:

1)

y=8x^4-5x^2-2/4x^3

(32x^3-10x)(4x^3)-(12x^2)(8x^4-5x^2-2)/(4x^3)^2

128x^6-40x^4+96x^6-60x^4-24x^2/(4x^3)^2

224x^6-100x^4-24x/16x^6

4x(56x^5-25x^3-24)/16x^6

y’=56x^5-25x^3-24/4x^5

2) y=sqroot 5x - sqroot x/5
y=(5x)^1/2 - (x/5)^1/2
y'=1/2(5x)^-1/2 (5) -1/2(x/5)^-1/2 (1/5)
y'=2/5(5x)^-1/2 -1/10(x/5)^-1/2
y'=2/5sqroot(5x) - 1/10(sqroot x/5)

Last edited: Jun 21, 2006
2. Jun 21, 2006

### arildno

1. You should use parentheses in order to indecate numerators and denominator in an unamiguous manner.

3. In your third line, you've forgotten to change signs when removing a parenthesis

3. Jun 21, 2006

### Hollysmoke

I typoed on the first one. Let me edit it. Thanks.

4. Jun 21, 2006

### arildno

5. Jun 21, 2006

### Hollysmoke

Okay I got the 1st one now for sure. Is the 2nd one correct?

6. Jun 21, 2006

### Hollysmoke

Okay one more question:

2x^3+2y^3-9xy=0

dy/dx = 9y+6x^2/9x-6y^2?

7. Jun 21, 2006

### HallsofIvy

You were just "bawled out" for not using parentheses!

No, the dy/dx is NOT (9y+ 6x^2)/(9x- 6y^2). You' missed a negative signs.

8. Jun 22, 2006

### VietDao29

You are wrong when going from the former line to the latter one.
$$\frac{5}{2 \sqrt{5x}} - \frac{1}{10 \sqrt{\frac{x}{5}}}$$
It's 5/2, not 2/5. :)
There's another way to do the first problem. Divide all by the denominator to get:
$$y = \frac{8x ^ 4 - 5 x ^ 2 - 2}{4x ^ 3} = 2x - \frac{5}{4x} - \frac{1}{2x ^ 3}$$
Now let's differentiate it with respect to x, we have:
$$y' = 2 + \frac{5}{4x ^ 2} + \frac{3}{2x ^ 4}$$
For the last problem, as HOI pointed out, you've missed quite a few negative signs. :)

9. Jun 22, 2006

### Hollysmoke

Thank you very much for your help! I found my errors now and found out what I did wrong. Thanks.

Given f(x) = x^3-0.5x^2+1, find the value of x where dy/dx = 4

So I found the derivative: f'(x)=3x^2-x
Tried setting it equal to 4

4=3x^2-x

But where do I go from there? Do I rearrange it and use quadratic formula to find the value of x?

10. Jun 22, 2006

### GregA

that should work

11. Jun 22, 2006

### Hootenanny

Staff Emeritus
You could do that but an easier option would be to factorise;

$$3x^2 - x - 4 = (3x-4)(x+1) = 0$$

12. Jun 22, 2006

### HallsofIvy

"HOI"??

I have a friend who uses the ename "Hog on Ice" which he abbreviates as HOI!

13. Jun 23, 2006

### Hootenanny

Staff Emeritus
I've heard of Bamby on Ice, but never "Hog on Ice" :rofl: :rofl: . A quick google didn't reveal its meaning, perhaps you could enlighten us HOI?