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