First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

Flop21 · May 1, 2016

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)

turntaker · May 1, 2016

Re: MATH1131 help thread

Flop21 said:
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

Flop21 · May 1, 2016

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

parad0xica · May 1, 2016

Re: MATH1131 help thread

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

Flop21 · May 1, 2016

Re: MATH1131 help thread

turntaker said:
Did you keep re applying the rule

Yeah and I got 6 not 3

Flop21 · May 1, 2016

Re: MATH1131 help thread

parad0xica said:
$\lim_{x \to 1} \frac{2x^4 - 3x^3 + x}{(x-1)^2} \stackrel{L'H}{=} \lim_{x \to 1} \frac{8x^3 - 9x^2 + 1}{2(x-1)} \stackrel{L'H}{=} \lim_{x \to 1} \frac{24x^2 - 18x}{2} \stackrel{x = 1}{=} 3$

Oh right thanks, yeah my mistake was silly, missing out 2 on the bottom.

parad0xica · May 1, 2016

Re: MATH1131 help thread

Flop21 said:
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?

iforgotmyname · May 1, 2016

Re: MATH1131 help thread

Flop21 said:
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

Flop21 · May 1, 2016

Re: MATH1131 help thread

parad0xica said:
With respect x?

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

parad0xica · May 1, 2016

Re: MATH1131 help thread

Flop21 said:
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,

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

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

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

Flop21 · May 1, 2016

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^?

InteGrand · May 1, 2016

Re: MATH1131 help thread

Flop21 said:
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 $y^{\prime} = \frac{1}{2\sqrt{x-1}}$.$

Flop21 · May 1, 2016

Re: MATH1131 help thread

InteGrand said:
$\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.

InteGrand · May 1, 2016

Re: MATH1131 help thread

Flop21 said:
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.

Flop21 · May 1, 2016

Re: MATH1131 help thread

InteGrand said:
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?

InteGrand · May 1, 2016

Re: MATH1131 help thread

Flop21 said:
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 $\frac{\mathrm{d}}{\mathrm{d} x}\left( \sqrt{x-1}\right)= \frac{1}{2\sqrt{x-1}}$.$

Flop21 · May 1, 2016

Re: MATH1131 help thread

InteGrand said:
$\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.

Flop21 · May 8, 2016

Re: MATH1131 help thread

How do I solve this with MAPLE?

Flop21 · May 8, 2016

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??

Flop21 · May 8, 2016

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?

First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

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