# Homework Help: Derivatives with pictures (problems 3-4)

1. Nov 20, 2008

### asdfsystema

Help me out please ! for #4 I'm preeeetty sure i did it correctly but it wouldn't hurt having someone double check it. for #5 , I'm not sure if i took the second derivative correctly

Will be on for the next couple hours trying to understand calculus ... :(

2. Nov 20, 2008

### Dick

You are pretty good on 4), the other two are an absolute disaster. I don't think you quite get the chain rule. Let's just take a part. What's d/dx(x*y)? Use the product rule also.

3. Nov 20, 2008

### asdfsystema

d/dx(xy) = (y)+(x)(dy/dx)

but i dont see where im supposed to apply the chain rule
i'm having trouble with the second part...

d/dx(xy^3)= (y^3) + (3y^2)*(dy/dx)*x

d/dx(10)=0

then I got [y^3+x*3y^2*dy/dx] + [y+x*dy/dx] = 0

EDIT: wait is d/dx (xy^3) = y^3+x3y *dy/dx

4. Nov 20, 2008

### Dick

Now you've got it right. Good job. That's different from what you wrote in the post. Now what is dy/dx?

5. Nov 20, 2008

### asdfsystema

[(y^3+x3y*dy/dx)] = [-(y+x*dy/dx)]

how do i cancel the dy/dx ?

since i'm almost done, i'll try to work on the other problem too. btw, is my #4 correct?

#5 f(x)= -2ln[sin(x)]

f'(x)= -2cos(x)/sin(x) --> -2cos(x)/sin(x) . b/c ln(u(x))' = u'(x)/u(x)

f"(x)= quotient rule ? sin(x)*2sin(x)+cos(x)(2cos(x)) / sin(x)^2

Thanks a lot ! ^^

6. Nov 20, 2008

### Dick

You don't cancel dy/dx. You solve for it. Move all the dy/dx parts to one side of the equation, factor dy/dx out and move everything else to the other side. And, yes, I think you are doing ok with part 4 as well. You could also say f'(x)=-2*cot(x) and go from there.

7. Nov 20, 2008

### asdfsystema

Haha I think I'm doing this wrong but ...

#3. -y^3-y / x3y / x and then i have to use tangent line equation after this. but i'll do it after you confirm its correct or wrong

f"(x) = -2csc^(x) is the answer ?

8. Nov 20, 2008

### Dick

I get +2csc(x)^2, but you might want to check that, it's late here. For the first one, the numerator looks good, but what's going on the denominator? x3y?

Last edited: Nov 21, 2008
9. Nov 21, 2008

### asdfsystema

Yup, you were correct but it's f"(x) 2csc^2(x)

and for this one...

Wait, was the one in the "EDIT" correct or incorrect?

but if its correct, -y^3 -y / 3y after cancelling out the x's

10. Nov 21, 2008

### Dick

You can't cancel that x, now can you?

11. Nov 21, 2008

### asdfsystema

i dont understand this at all . its confusing me , i'm not even sure if i'm supposed to look at

1. d/dx (xy^3) = y^3+x3y *dy/dx

or

2.d/dx(xy^3)= (y^3) + (3y^2)*(dy/dx)*x

assuming its 1. this time i got (-y^3-y)(x) / x3y .............

after i use the implicit differentiation, to find the tangent line to the curve , i use point slope formula right ?

12. Nov 21, 2008

### Dick

It's 2. d/dx(y^3)=3y^2*dy/dx. Chain rule. You had it right when you said "I got [y^3+x*3y^2*dy/dx] + [y+x*dy/dx] = 0". Solve that for dy/dx. I'll repeat what I already said. "Move all the dy/dx parts to one side of the equation, factor dy/dx out and move everything else to the other side." By dividing by it. Like this: if x+y=x*dy/dx-y*dy/dx, dy/dx=(x+y)/(x-y).

Last edited: Nov 21, 2008
13. Nov 21, 2008

### asdfsystema

okay this time i got -y^3-y / 3y^2 . damn i hope this is right .

with all these "y's" how can i use the point slope formula to get y=mx+b for question #3 (the one dealing with implicit differentiation)

14. Nov 21, 2008

### Dick

Look at your correct differentiation in the preceding post. Where did the term x*dy/dx go? And what happened to the x in x*3*y^2? You'll finally use (x,y)=(5,1) and figure out dy/dx which is m in your point slope form. Can you show exactly how you solved for dy/dx?

15. Nov 21, 2008

### asdfsystema

okay i am getting really confused .

here's what i did so far :

I thought you told me to just leave the 2 dy/dx's on the left side and then I could cancel out the x if i bring it to the right side x/x = 1

16. Nov 21, 2008

### Dick

Take your second line from the bottom. That becomes (x*3y^2+x)*dy/dx=(-y^3-y). So dy/dx=(-y^3-y)/(x*3y^2+x). You can't move the 3y^2 over all by itself, and you can't just 'cancel' the x. There's nothing to 'cancel' it. Those aren't calculus problems, those are algebra problems.

17. Nov 21, 2008

### asdfsystema

Thanks. the next thing I have to do is find the tangent line to this...

i need to find the slope first. Do i need to take derivative again? I don't think so because i already did implicit differentiation...

so the next step is to plug in the x value (5,1) so i plug in 5 ... into all the x's.

I haev a question, do I get y alone on one side of the equation and then plug in the x ?

18. Nov 21, 2008

### Dick

Yes, find the slope first. That's dy/dx. You have a formula for dy/dx. Put x=5 and y=1 into it.

19. Nov 21, 2008

### asdfsystema

after plugging in, i get dy/dx = -1/10 which is the slope

then i use point slope formula y-y0 = m (x -x0)

y-(1) = -1/10x-1/2

i get y= -1/10x + 1/2 so m= -1/10 and b= 1/2 ?

20. Nov 21, 2008

### Dick

Your problems with algebra are killing you. -1/10*(x-5)=(-1/10)*x+1/2 NOT (-1/10)*x-1/2. You should also practice checking yourself. If you think y=(-1/10)*x+1/2, check it. If you put in x=5, you should get y=1. You don't.

Last edited: Nov 21, 2008