# Thread: First Year Mathematics A (Differentiation & Linear Algebra)

1. ## Re: MATH1131 help thread

Can someone do this via l'hopital's rule? Answer is 3, which I'm not getting for some odd reason.

lim x -> 1 (2x^4-3x^3+x)/((x-1)^2)

2. ## Re: MATH1131 help thread

Originally Posted by Flop21
Can someone do this via l'hopital's rule? Answer is 3, which I'm not getting for some odd reason.

lim x -> 1 (2x^4-3x^3+x)/((x-1)^2)
Did you keep re applying the rule

3. ## Re: MATH1131 help thread

Can someone also do this question, I'm not getting the right answer.

Differentiate x + ln(x) = y + 2ln(y)

I keep getting 1 + 1/x - 2/y

4. ## Re: MATH1131 help thread

$\lim_{x \to 1} \frac{2x^4 - 3x^3 + x}{(x-1)^2} \stackrel{L'H}{=} \lim_{x \to 1} \frac{8x^3 - 9x^2 + 1}{2(x-1)} \stackrel{L'H}{=} \lim_{x \to 1} \frac{24x^2 - 18x}{2} \stackrel{x = 1}{=} 3$

5. ## Re: MATH1131 help thread

Originally Posted by turntaker
Did you keep re applying the rule
Yeah and I got 6 not 3

6. ## Re: MATH1131 help thread

$\lim_{x \to 1} \frac{2x^4 - 3x^3 + x}{(x-1)^2} \stackrel{L'H}{=} \lim_{x \to 1} \frac{8x^3 - 9x^2 + 1}{2(x-1)} \stackrel{L'H}{=} \lim_{x \to 1} \frac{24x^2 - 18x}{2} \stackrel{x = 1}{=} 3$
Oh right thanks, yeah my mistake was silly, missing out 2 on the bottom.

7. ## Re: MATH1131 help thread

Originally Posted by Flop21
Can someone also do this question, I'm not getting the right answer.

Differentiate x + ln(x) = y + 2ln(y)

I keep getting 1 + 1/x - 2/y
With respect x?

8. ## Re: MATH1131 help thread

Originally Posted by Flop21
Can someone also do this question, I'm not getting the right answer.

Differentiate x + ln(x) = y + 2ln(y)

I keep getting 1 + 1/x - 2/y
You forgot about the chain rule at the end. And f'x of y is dy/dx

9. ## Re: MATH1131 help thread

With respect x?
Uhhhh not sure. It's implicit differentiation sorry. It's part of a equation of a line tangent question.

10. ## Re: MATH1131 help thread

Originally Posted by Flop21
Uhhhh not sure. It's implicit differentiation sorry. It's part of a equation of a line tangent question.
To demonstrate iforgotmyname's comment:

$x + \ln(x) = y + 2 \ln(y)$

Differentiate all terms with respect to x,

$\frac{d}{dx}x + \frac{d}{dx} \ln(x) = \frac{d}{dx}y + 2 \frac{d}{dx}\ln(y)$

$1 + \frac{1}{x} = \frac{dy}{dx} + 2*\frac{1}{y}*\frac{dy}{dx}$

$\frac{dy}{dx} = \dots$

11. ## Re: MATH1131 help thread

Why can't I used chain rule to differentiate sqrt(x-1), e.g. (x-1)^1/2... y'= 1/2(x-1)*1

Instead the answers use the definition of a derivative to solve and get a different answer. Am I doing something wrong above^?

12. ## Re: MATH1131 help thread

Originally Posted by Flop21
Why can't I used chain rule to differentiate sqrt(x-1), e.g. (x-1)^1/2... y'= 1/2(x-1)*1

Instead the answers use the definition of a derivative to solve and get a different answer. Am I doing something wrong above^?
$\noindent What is the full question? If the answers use the definition, maybe the question has something in it requiring us to do so. Incidentally, if y=\sqrt{x-1}, then y^{\prime} = \frac{1}{2\sqrt{x-1}}.$

13. ## Re: MATH1131 help thread

Originally Posted by InteGrand
$\noindent What is the full question? If the answers use the definition, maybe the question has something in it requiring us to do so. Incidentally, if y=\sqrt{x-1}, then y^{\prime} = \frac{1}{2\sqrt{x-1}}.$
So you get different derivatives using normal methods vs the definition?

The question is a Mean Value Theorem question, with that function on an interval.

14. ## Re: MATH1131 help thread

Originally Posted by Flop21
So you get different derivatives using normal methods vs the definition?

The question is a Mean Value Theorem question, with that function on an interval.
We should get the same answer.

I can't see the specific Q., but if it's that function over an interval (assuming it's well-defined there, i.e. x > 1 there), then the derivative should be what I posted.

15. ## Re: MATH1131 help thread

Originally Posted by InteGrand
We should get the same answer.

I can't see the specific Q., but if it's that function over an interval (assuming it's well-defined there, i.e. x > 1 there), then the derivative should be what I posted.
It is, but what I'm I doing incorrect? Or how are you getting that derivative?

16. ## Re: MATH1131 help thread

Originally Posted by Flop21
It is, but what I'm I doing incorrect? Or how are you getting that derivative?
$\noindent It's just a standard power rule. If y=\sqrt{x-1}, then y=\left(x-1\right)^{\frac{1}{2}}. So y^{\prime} = \frac{1}{2}\left(x-1\right)^{\frac{1}{2}-1}= \frac{1}{2}\left(x-1\right)^{-\frac{1}{2}}=\frac{1}{2\left(x-1\right)^{\frac{1}{2}}}=\frac{1}{2\sqrt{x-1}}.$

$\noindent It's worth remembering the derivative of the square root function: \frac{\mathrm{d}}{\mathrm{d}x} \left( \sqrt{x}\right)=\frac{1}{2\sqrt{x}}. The reason it's worth remembering is that square roots come up quite a lot. It follows from this and the chain rule that \frac{\mathrm{d}}{\mathrm{d} x}\left( \sqrt{x-1}\right)= \frac{1}{2\sqrt{x-1}}.$

17. ## Re: MATH1131 help thread

Originally Posted by InteGrand
$\noindent It's just a standard power rule. If y=\sqrt{x-1}, then y=\left(x-1\right)^{\frac{1}{2}}. So y^{\prime} = \frac{1}{2}\left(x-1\right)^{\frac{1}{2}-1}= \frac{1}{2}\left(x-1\right)^{-\frac{1}{2}}=\frac{1}{2\left(x-1\right)^{\frac{1}{2}}}=\frac{1}{2\sqrt{x-1}}.$
Oh my. I was forgetting to do the 0.5-1 to the power.

Thanks appreciate it InteGrand.

18. ## Re: MATH1131 help thread

How do I solve this with MAPLE?

19. ## Re: MATH1131 help thread

How do I do this in MAPLE?

I understand you use sum(); but what do you do with the n^5/2 on the left??

20. ## Re: MATH1131 help thread

How do I solve this in MAPLE?

My incorrect attempt was to go... s:= [solve(expression)] then allvalues(s[1]); to find first root... same for all other roots, then evalf(result, 10); on each root.

Anyone know the steps to the correct way to get right answers?

21. ## Re: MATH1131 help thread

Originally Posted by Flop21
How do I do this in MAPLE?

I understand you use sum(); but what do you do with the n^5/2 on the left??
The expression in the limit is not a sum, but rather a product. This is because that symbol is capital Pi ($\Pi$), which is used for products. The symbol for summation is a capital Sigma ($\Sigma$).

22. ## Re: MATH1131 help thread

Originally Posted by InteGrand
The expression in the limit is not a sum, but rather a product. This is because that symbol is capital Pi ($\Pi$), which is used for products. The symbol for summation is a capital Sigma ($\Sigma$).
yeah sorry I actually meant product(); originally.

23. ## Re: MATH1131 help thread

bump someone help me with these MAPLE questions pls

24. ## Re: MATH1131 help thread

Originally Posted by Flop21
How do I do this in MAPLE?

I understand you use sum(); but what do you do with the n^5/2 on the left??
Try

limit(n^(5/2)*(product(2*k/(2*k+5), k = 1 .. n)), n = infinity)

25. ## Re: MATH1131 help thread

Originally Posted by Flop21
How do I solve this in MAPLE?

My incorrect attempt was to go... s:= [solve(expression)] then allvalues(s[1]); to find first root... same for all other roots, then evalf(result, 10); on each root.

Anyone know the steps to the correct way to get right answers?
Close, but maple can calculate all roots for a polynomial like that at once, so you don't need to specify an interval. Your first part, solve z for p(z)=0 is correct. This doesn't give numerical values though, so you need to use evalf.

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