Help with differentiation question (1 Viewer)

girlworld_club · Jul 8, 2013

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2003exams/pdf_doc/mathematics_03.pdf

question 10 part ii)

i just don't seem to get the right answer, probably because my differentiation has become quite a bit pitchy over the last year.

HeroicPandas · Jul 8, 2013

girlworld_club said:
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2003exams/pdf_doc/mathematics_03.pdf

question 10 part ii)

i just don't seem to get the right answer, probably because my differentiation has become quite a bit pitchy over the last year.

$l = 2 + (r^2 - x^2)^{\frac{1}{2}}- (1-2x + r^2)^{\frac{1}{2}} \\ \\ \frac{dl}{dx} = -x(r^2 - x^2)^{\frac{-1}{2}} + (1-2x + r^2)^{\frac{-1}{2}}\\\\ \\ \frac{dl}{dx} = \frac{-x}{\sqrt{r^2 - x^2}} + \frac{1}{\sqrt{1-2x+r^2}} \\\\\\ \frac{dl}{dx} = \frac{-x\sqrt{1-2x+r^2}+\sqrt{r^2 - x^2}}{\sqrt{r^2 - x^2}\sqrt{1-2x + r^2}} \\\\\\ \frac{dl}{dx} = \frac{-x\sqrt{1-2x+r^2}+\sqrt{r^2 - x^2}}{\sqrt{r^2 - x^2}\sqrt{1-2x + r^2}} \cdot \frac{\sqrt{r^2 - x^2}+ x\sqrt{1-2x+r^2}}{\sqrt{r^2 - x^2}+ x\sqrt{1-2x+r^2}} \\\\ \\ \frac{dl}{dx} = \frac{(r^2 - x^2) - x^2 (1-2x+r^2)}{\sqrt{r^2 - x^2}\sqrt{1-2x+r^2}\left ( \sqrt{r^2 - x^2}+ x\sqrt{1-2x+r^2} \right )}$

I used differentiation, chain rule, common denomination, rationalisation (accompanied with logic) and difference of two squares (but i didnt indicate where - ur job to find them)

girlworld_club · Jul 8, 2013

thank you sooo much for that! it has cleared a lot of things up. However, just wondering how you got from line 1, to line 2. if anyone else is willing to help, you've already done a lot.

HeroicPandas · Jul 8, 2013

$\alpha = (r^2 - x^2)^{\frac{1}{2}} \\ \\ \therefore \frac{d\alpha}{dx} = \frac{1}{2}(r^2 - x^2)^{\frac{1}{2}-1}\left ( \frac{d(-x^2)}{dx} \right ) \\ \\ \therefore \frac{d\alpha}{dx} = \frac{1}{2}(r^2 - x^2)^{\frac{-1}{2}} (-2x) \\ \\ \therefore \frac{d\alpha}{dx} = -x(r^2 - x^2)^{\frac{-1}{2}}$

Remember to differentiate the INSIDE

omgiloverice · Jul 8, 2013

$y & = & (r^{2}-x^{2})^{\frac{1}{2}}\\ \\ Let\ u & = & r^{2}-x^{2}\\ \\ \frac{du}{dx} & = & -2x\\ \\ \frac{dy}{du} & = & \frac{1}{2}u^{-\frac{1}{2}}\\ \\ \frac{dy}{dx} & = & \frac{dy}{du}\cdot\frac{du}{dx}\\ \\ & = & -x(u)^{-\frac{1}{2}}\\ \\ & = & -x(r^{2}-x^{2})^{-\frac{1}{2}}$

Chain rule

girlworld_club · Jul 8, 2013

isen't the diff of (r^2 - x^2) = 2r - 2x

why do you only take into account the -2x

omgiloverice · Jul 8, 2013

girlworld_club said:
isen't the diff of (r^2 - x^2) = 2r - 2x

why do you only take into account the -2x

'r' is a constant so treat it like an ordinary number. It is like differentiating f(x)=(2^2 -x^2)

asianese · Jul 9, 2013

When you are differentiating with respect to x ONLY, r^2 is merely a constant. Hence its derivative is 0, and the derivative of -x^2 with respect to x is -2x.

girlworld_club · Jul 9, 2013

so if we had 2y - 2x^2 and we were diff in respect to x it would still be -2x right?

and thanks.

girlworld_club · Jul 9, 2013

but where in the question does it say diff in respect to x???

SpiralFlex · Jul 9, 2013

girlworld_club said:
so if we had 2y - 2x^2 and we were diff in respect to x it would still be -2x right?

and thanks.

-4x [assuming y is constant]

SpiralFlex · Jul 9, 2013

girlworld_club said:
but where in the question does it say diff in respect to x???

Very valid point. However most 2u papers and texts will imply this. Only in 3/4u you may come across something called implicit differentiation. Though correctly they should say it

Drongoski · Jul 9, 2013

girlworld_club said:
so if we had 2y - 2x^2 and we were diff in respect to x it would still be -2x right?

and thanks.

If you differentiate 2y - 2x^2 with respect to x you get:

$2\frac {dy}{dx} - 4x$

since 2y is not a constant term.

asianese · Jul 9, 2013

That's only if you consider y as a function of x, which is rare in 2u

girlworld_club · Jul 9, 2013

thank you everyone!! This is the first time i have actually encountered something like this. many thanks

girlworld_club · Jul 9, 2013

I also have another question: How would you differentiate:

x^2 + 16 loge (5-x)^2

i got 2x (-32)/(5-x)

but the answer is:

2x (-32loge (5-x)) / (x-5)

why do they keep the loge?

HeroicPandas · Jul 9, 2013

girlworld_club said:
I also have another question: How would you differentiate:

x^2 + 16 loge (5-x)^2

i got 2x (-32)/(5-x)

but the answer is:

2x (-32loge (5-x)) / (x-5)

why do they keep the loge?

keep in mind, i do extra steps so u can see the light
$\alpha= x^2 + 16\left ( ln(5-x)\right)^2 \\ \\ \frac{d\alpha}{dx} = 2x + (2)(16)ln(5-x)\cdot \frac{d(ln(5-x))}{dx} \\ \\ \frac{d\alpha}{dx} = 2x + (32ln(5-x))\cdot \frac{-1}{5-x}$

FOR THE BOLDED PART - can u show me ur working out?

girlworld_club · Jul 9, 2013

this is the actual working for the q:

y = x^2 + 16 loge (5-x)^2
y' = 2x + 16 . 2 (loge 5-x) . -1/5-x

where did u go wrong?

girlworld_club · Jul 9, 2013

for some reason i also don't get the answer to this simple probability question:

There are two groups of people at a part and Minh is blind folded. In the first group there are 4 men, 3 women and 2 children. In the second there are 7 men and 5 women. Minh is spun and asked to select one person at random.

i ) find the probability that minh approaches a person in the first group and then selects a women.
ii)find the probability that a women from either group is selected.

HeroicPandas · Jul 9, 2013

girlworld_club said:
for some reason i also don't get the answer to this simple probability question:

There are two groups of people at a part and Minh is blind folded. In the first group there are 4 men, 3 women and 2 children. In the second there are 7 men and 5 women. Minh is spun and asked to select one person at random.

i ) find the probability that minh approaches a person in the first group and then selects a women.
ii)find the probability that a women from either group is selected.

(i)
Prob of choosing FIRST group is 1/2
Prob of choosing a woman in the first group is 3/9

Prob (first group, woman chosen) = 1/2 x 3/9 = 1/6

(ii)
Prob of choosing SECOND group = 1/2
Prob of choosing a lady in second group = 5/12

Prob (second group, woman chosen) = 1/2 x 5/12 = 5/24

Total Prob (picking a woman = 5/24 + 1/6 = 9/24 = 3/8

Is this correct or wrong?

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