C3 Diff. Products - I'm so confused as to what I am doing wrong. Watch

sham91
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Textbook: Cambridge Advanced Mathematics Core 3 & 4 by Hugh Neil and Douglas Quadling

Chapter 10 Diff Products: Ex 10B page 171.

Question 5.

Find the equation of the tangent at the point with coordinates (1,1) to the curve with equation y= (x^2 +3)/(x + 3)


From my understanding your supposed to use the quotient rule which I have been doing and my final answer is always coming out to be 8y = 5x + 3

however in the answers it says the answer is 4y = x + 3


Here is my working:

y= (x^2+ 3)/(x + 3)

u = (x^2 + 3) v = (x + 3)

du/dx = 2x dv/dx = 1


Then using quotient rule: ((du/dx )(v) - (u)(dv/dx)) / v^2


(2x)(x+3) - (x^2 + 3)(1) / (x+3)^2 = (x^2 + 6x + 3) / (x+3)^2 = gradient of tangent


and when x = 1 the gradient of tangent = 1^2 + 6(1) + 3 / (1+3)^2 = 5/8


then using y-y1 = m(x-x1)


y - 1 = 5/8(x - 1) expanding and simplifying = 8y=5x + 3


and the answer in the back of the book says it is supposed to be 4y = x + 3








Can somebody do a quick step by step please? Many thanks and i'll give rep+ and all that good stuff.
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nuodai
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(Original post by sham91)
From my understanding your supposed to use the quotient rule which I have been doing and my final answer is always coming out to be 8y = 5x + 3

however in the answers it says the answer is 4y = x + 3

Can somebody do a quick step by step please? Many thanks and i'll give rep+ and all that good stuff.
Why not show your working? Then we can tell you where you're going wrong.
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dknt
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(Original post by sham91)
Textbook: Cambridge Advanced Mathematics Core 3 & 4 by Hugh Neil and Douglas Quadling

Chapter 10 Diff Products: Ex 10B page 171.

Question 5.

Find the equation of the tangent at the point with coordinates (1,1) to the curve with equation y= (x^2 +3)/(x + 3)


From my understanding your supposed to use the quotient rule which I have been doing and my final answer is always coming out to be 8y = 5x + 3

however in the answers it says the answer is 4y = x + 3

Can somebody do a quick step by step please? Many thanks and i'll give rep+ and all that good stuff.
Post your working and we can see where you have gone wrong. (If you have gone wrong that is. I haven't checked.)
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nuodai
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(Original post by dknt)
(If you have gone wrong that is. I haven't checked.)
They have
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Miss Mary
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It's ike you've said OP, you have to use the quotient rule:
f(x)=\frac{g(x)}{h(x)}
f'(x)=\frac{h(x)g'(x)-h'(x)g(x)}{[h(x)]^2}

Make sure you're not forgetting the minus sign
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dknt
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(Original post by sham91)
(2x)(x+3) - (x^2 + 3)(1) / (x+3)^2 = (x^2 + 6x + 3) / (x+3)^2 = gradient of tangent
You sure about that?
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sham91
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(Original post by dknt)
You sure about that?


yeah i am sure it is +3. isn't it?

(2x)(x+3) - (x^2 + 3)(1) = (2x^2 + 6x) - (x^2 + 3) = x^2 + 6x + 3
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dknt
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(Original post by sham91)
yeah i am sure it is +3. isn't it?

(2x)(x+3) - (x^2 + 3)(1) = (2x^2 + 6x) - (x^2 + 3) = x^2 + 6x + 3
-(+3)=+3?
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sham91
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(Original post by dknt)
-(+3)=+3?
lol. im blind. i'll try that and see if get the right answer.
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ok_cub2008
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First advice... **** THE QUOTIENT RULE.

Real gangsters write the denominator in the numerator as power to the -1, then use product rule *****.

Eaaaasyyyy.

Next week on this show, NEVER USE THE QUADRATIC FORMULA.

Real gangsters complete the square, you dig?
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sham91
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(Original post by sham91)
lol. im blind. i'll try that and see if get the right answer.
Sir, you have the eye of an eagle. Thank you. It turns out that's what I was doing wrong. Thank you very much.
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dknt
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(Original post by sham91)
Sir, you have the eye of an eagle. Thank you. It turns out that's what I was doing wrong. Thank you very much.
Glad I could help
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ok_cub2008
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I don't really get all the negative rep on my comment. I seriously think you should consider throwing the quotient rule out the window. You might forget it, or just get it confused. The key to doing well is keeping it simple, and not relying on memorised formulas without understanding what is going on.
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nuodai
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(Original post by ok_cub2008)
I don't really get all the negative rep on my comment. I seriously think you should consider throwing the quotient rule out the window. You might forget it, or just get it confused. The key to doing well is keeping it simple, and not relying on memorised formulas without understanding what is going on.
Judging by the timestamp on this post and the neg-rep I just received I'm guessing that was from you... but for what it's worth, I never negged you. In fact, I agree with your sentiment. I think the quotient rule and quadratic formula are worth memorising though, provided you know how to derive them too, because they're very useful.
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Dirac Spinor
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I didn't neg you but I disagree with you. the quadratic formula in particular has it's place. Even in STEP questions it is an invaluable tool.
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ok_cub2008
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(Original post by nuodai)
Judging by the timestamp on this post and the neg-rep I just received I'm guessing that was from you... but for what it's worth, I never negged you. In fact, I agree with your sentiment. I think the quotient rule and quadratic formula are worth memorising though, provided you know how to derive them too, because they're very useful.
"Providing you know how to derive them" is the key!!
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nuodai
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(Original post by ok_cub2008)
"Providing you know how to derive them" is the key!!
Well sort of. You can use the chain and product rules to derive the quotient rule by writing \dfrac{f(x)}{g(x)} = f(x)g(x)^{-1}, but can you derive the product rule? Can you derive the chain rule? In fact, before even proceeding you should be able to check if f,g are both differentiable and we should calculate their derivatives from first principles. I mean, why would you just quote the result \dfrac{d}{dx} (x^n) = nx^{n-1} when you can derive it from first principles using the binomial theorem? After all, it's very easy to screw up if you're just memorising formulae. But whilst we're at it, we should really derive the binomial theorem. I suppose we can do this by induction, but it seems silly to use induction if we haven't defined what the integers are, so we'd better get Peano on this question's ass.

Sarcasm aside, it's very difficult to know where to draw the line with things like this, so it seems silly to say that you should never memorise certain things unless you know how to derive them.
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ok_cub2008
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(Original post by nuodai)
Well sort of. You can use the chain and product rules to derive the quotient rule by writing \dfrac{f(x)}{g(x)} = f(x)g(x)^{-1}, but can you derive the product rule? Can you derive the chain rule? In fact, before even proceeding you should be able to check if f,g are both differentiable and we should calculate their derivatives from first principles. I mean, why would you just quote the result \dfrac{d}{dx} (x^n) = nx^{n-1} when you can derive it from first principles using the binomial theorem? After all, it's very easy to screw up if you're just memorising formulae. But whilst we're at it, we should really derive the binomial theorem. I suppose we can do this by induction, but it seems silly to use induction if we haven't defined what the integers are, so we'd better get Peano on this question's ass.

Sarcasm aside, it's very difficult to know where to draw the line with things like this, so it seems silly to say that you should never memorise certain things unless you know how to derive them.
My point is you have to keep it simple, use as few separate rules as possible and try to have some central logic that you work from. I can't derive the simple rule for taking the derivative of a function of x, so I take it as given.

But you have a good point about product rule, I haven't a clue how to derive that. But knowing product rule, I can "work out" stuff like the quotient rule and the formula for integration by parts, rather than knowing all 3 separately.

I think people that are naturally good at maths (at least in terms of annihilating A-level standard exams) tend to do this as second nature.
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