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Calculus Help............
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I'm doing a worksheet and these are the last 2 and I'm really tired and don't know how to do them. I know it's a stretch considering most of NS is 12 and semi-retarded, but I thought someone might know. Thanks. We're doing derivitives by the way.
1. What is the derivative of y=x+sin(xy) in terms of x and y, at any point on the curve?
2. f(g(x))=x, f(2)=7, and f'(2)=6. What is g'(7)?
If you can get any of these, thanks, if not, I'm gonna have to go ask someone smarter
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I'd help but I am in a different class, and it'd be hard to explain.
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The deriative of sin is -cos...if I remember correctly. Its been 2 years I havent done any calculus.
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iam ghlad iam dun whith thaht shite. Fhamilie and Cunsumr Scynces, hell yea!
eyed sugjest ass-king moure entellighent phersons.
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1. What is the derivative of y=x+sin(xy) in terms of x and y, at any point on the curve?
So I'm assuming they want dy/dx, in which case you can use implicit differentiation:
(dy/dx)=1+cos(xy)[x(dy/dx)+y]
(dy/dx)=1 +xcos(xy)(dy/dx) +ycos(xy)
Group the (dy/dx) terms together
(dy/dx)-xcos(xy)(dy/dx)=1+ycos(xy)
Factor out (dy/dx)
(dy/dx)[1-xcos(xy)]=1+ycos(xy)
Answer:
(dy/dx)=(1+ycos(xy))/(1-xcos(xy))
2. f(g(x))=x, f(2)=7, and f'(2)=6. What is g'(7)?
I can't rememeber how to do this....What chapter are you on? I'll look it up for you.
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No idea if this is right...buuut...
2) f(g(x))=x, f(2)=7, f'(2)=6
f'(g(x))=f'(u) X g'(x) u=g(x)
f'(2)=f'(2) X g'(x)=6
f'(2)=6 X g'(x)=6
g'(x) = 6/6 = 1
g'(x) = 1
g'(x) = x
Integral of g'(x) = g(x) = (1 X x^2)/2 = (x^2)/2
g(x) = (x^2)/2
g'(x) = (2x)/2 = x
g'(70 = 7
Like I said, not sure if thats right...its been awhile since I've had calc. Its also slightly redundant but i did that to make it seem more sensible. Still working on the first one.
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g'(7) = 7, my bad, hit 0 not )
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g'(x) = 1
g'(x) = x
How did you make that assumption?
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Hmmm there was a reason but I forgot it haha. Back to the books!
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Haha, I can't find it anywhere in my book. I completely failed that part back in calc 1. I didn't even try a single problem like it. I don't even know what it's called. I thought it would be like "composite functions" or something, but I couldn't find the same type of problem in that section.
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Yaaaa I just got done with an exam at 7:45 this morning and I have another coming up, so dead right now haha. Looking back at that, it doesn't really make sense. But I think the g'(x) = 1 part is still correct. This means that you g'(7) = 1...hopefully haha I dunno, I give up lol.
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I'm studying for a Calc 3 exam, I don't have time for petty calc 1. Haha. #1 should be right though, assuming it was asking for dy/dx and implicit differentiation.
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I'm pretty sure it was. I was about to do it...and then I saw you did and I was happy haha.
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K, I'm going to go to my calc 3 class now. Nice working with you, haha.
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Holy fucking shit!!!!!!!!!!!!!!!!! I jsut typed this huge ass fucking response to the 2nd problem and hit back on accident and it deleted everything i just typed!!!!!!!!!!!!!!!!! Wow...I am so pissed off. I spent 20 minutes typing it.
The answer I got was 1/6 after finding f'(x) and then integrating to get f(x). I substituted g(x) into f(x) as such:
I found f(x) to be (3/2)x^2+1
Substituted g(x) so (3/2)(g(x))^2+1
Set that equal to f(g(x))=x
Solved for g(x) and got sqroot[(2/3)(x-1)]
I checked this by plugging it into f(x) as f(sqroot[(2/3)(x-1)]) and it came out to x as given.
Found the derivative of that and got 1/(3sqroot[(2/3)(x-1)) which is g'(x)
So g'(7) = 1/(3sqroot[(2/3)(6)])
g'(7)= 1/(3sqroot[4])
g'(7)=1/6
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That's a very simplified version of what I typed. It was step by step. Gah, I'm still pissed. Let me know if that's the right answer. I remember there being a much easier way to do this.
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Hey, I just want to say thanks for the calc help. I just did it out with one of my brighter friends, and you were right. So thanks.
Ben
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and fuck, I meant to message that and not quote it. ahh well
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