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astroman · Oct 3, 2014

stuck on a question, getting confused :/

Differentiate, find the stationary point on the curve, giving the exact value.

astroman · Oct 3, 2014

$y=x \sqrt{x+1}$
$y{}'=x.(x+1)^{\frac{1}{2}}$
$y{}'=\frac{1}{2}(x+1)^{\frac{-1}{2}}$

is this right so far

faisalabdul16 · Oct 3, 2014

astroman said:
stuck on a question, getting confused :/

Differentiate, find the stationary point on the curve, giving the exact value.

What do you mean by giving exact value?
Does it want the x coordinate or the y coordinate or both?

You differentiate.
Stationary points occur when first derivative = 0.
Solve for x. That is your x value.
Sub into original equation and you get a y value.
Coordinates of stationary point is (x,y)
Not sure what the question means about value, it's vague.

astroman · Oct 3, 2014

the questions says exactly: differentiate $y=x\sqrt{x+1}$ . Hence find the stationary point on the curve, giving the exact value.

question from EX 2.1 - 14, MIF Ext1

mreditor16 · Oct 3, 2014

astroman said:
the questions says exactly: differentiate $y=x\sqrt{x+1}$ . Hence find the stationary point on the curve, giving the exact value.

question from EX 2.1 - 14, MIF Ext1

MIF

MIF

MIF

MIF

MIF

oh oh :/

astroman · Oct 3, 2014

mreditor16 said:
MIF

MIF

MIF

MIF

MIF

oh oh :/

its all i have atm, better then nothing :/

Drongoski · Oct 3, 2014

$y = f(x) = x\sqrt{x+1} = (x^3+x^2)^{\frac {1}{2}} \\ \\ \therefore f'(x) = \frac{1}{2}(x^3+x^2)^{-\frac{1}{2}}(3x^2+2x) = \frac {x(3x+2)}{2\sqrt{x^3 + x^2}} = \frac {3x+2}{2\sqrt{x+1}} \\ \\ = 0 $ at $ x = -\frac{2}{3}$

Now find the co-ords of the stationary point.

Smile12345 · Oct 3, 2014

astroman said:
$y=x \sqrt{x+1}$
$y{}'=x.(x+1)^{\frac{1}{2}}$
$y{}'=\frac{1}{2}(x+1)^{\frac{-1}{2}}$

is this right so far

$y{}=x.(x+1)^{\frac{1}{2}}$

astroman · Oct 3, 2014

Drongoski said:
$y = f(x) = x\sqrt{x+1} = (x^3+x^2)^{\frac {1}{2}}$

answer from textbook says,

$\frac{x}{2\sqrt{x+1}}+\sqrt{x+1}=\frac{3x+2}{2\sqrt{x+1}} ; \left ( \frac{-2}{3} \right, \frac{-2\sqrt{3}}{9})$

enigma_1 · Oct 3, 2014

astroman said:
the questions says exactly: differentiate $y=x\sqrt{x+1}$ . Hence find the stationary point on the curve, giving the exact value.

question from EX 2.1 - 14, MIF Ext1

u w295t m9??

faisalabdul16 · Oct 3, 2014

astroman said:
answer from textbook says,

$\frac{x}{2\sqrt{x+1}}+\sqrt{x+1}=\frac{3x+2}{2\sqrt{x+1}} ; \left ( \frac{-2}{3} \right, \frac{-2\sqrt{3}}{9})$

Yeah that's what i got.

Smile12345 · Oct 3, 2014

Use the basic product rule... y = vu' + uv' and you will get what the first part of the answer is.

Then you need to put the two over a common denominator and you will get the next part of the answer.

Then as user 'Faisalabdul16' said:
"Stationary points occur when first derivative = 0.
Solve for x. That is your x value.
Sub into original equation and you get a y value.
Coordinates of stationary point is (x,y)"

You should get the co-ordinate as given.

faisalabdul16 · Oct 3, 2014

$y = x\sqrt{x+1} \\ \frac{dy}{dx} = \frac{x}{2\sqrt{x+1}} + \sqrt{x+1} \\ \\ $Stationary points occur when first derivative = 0 \\ \\ Multiply everything by the denominator and you get \\ x + 2(x+1)= 0 \\\therefore x = \frac{-2}{3} \\ \\$

Sub x value into y and you'll get coordinates of the stationary point.

astroman · Oct 3, 2014

faisalabdul16 said:
$y = x\sqrt{x+1} \\ \frac{dy}{dx} = \frac{x}{2\sqrt{x+1}} + \sqrt{x+1} \\ \\ $Stationary points occur when first derivative = 0 \\ \\ Multiply everything by the denominator and you get \\ x + 2(x+1)= 0 \\\therefore x = \frac{-2}{3} \\ \\$

Sub x value into y and you'll get coordinates of the stationary point.

dannnnng, who taught you such pro LaTeX?

seventhroot · Oct 3, 2014

Damm, I wasn't inb4enigma

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