Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

How do you differentiate-

((a^2-a+0.5)^0.5)/3?

I try using the chain rule, but just end up with a mess. Do I ignore the /e when differentiating the top?

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Is that

(a^2 -a + 0.5)^\frac{0.5}{3}

? or

\frac{(a^2 -a + 0.5)^(\frac{1}{2})}{3}

OP

JumpingJonny

Is that

(a^2 -a + 0.5)^\frac{0.5}{3}

? or

\frac{(a^2 -a + 0.5)^0.5}{3}

The latter. It's everything divided by 3.

You can take out the 1/3 as a constant:

\frac{d}{dx} \frac{(a^2-a+0.5)^{1/2}}{3} = \frac {1}{3} \frac{d}{dx} (a^2-a+0.5)^{1/2}

Then use the chain rule with the substitution

u = a^2-a+0.5

EierVonSatan

You can take out the 1/3 as a constant:

\frac{d}{dx} \frac{(a^2+a+0.5)^{1/2}}{3} = \frac {1}{3} \frac{d}{dx} (a^2+a+0.5)^{1/2}

Then use the chain rule with the substitution

u = a^2+a+0.5

This, I had to go get dinner so was beaten to it...again :rolleyes:

:rolleyes:

OP

Should I get

(2a-1)/(6(a^2-1+0.5)^0.5)

??

Unparseable latex formula:

\dfrac{(a^2-a+0.5)^0^.^5}{3}

Unparseable latex formula:

u = (a^2-a+0.5)}

\dfrac{du}{dx} = 2a - 1

Unparseable latex formula:

\dfrac{dy}{du} = (\dfrac{1}{2})(\dfrac{1}{3})u^-^0^.^5

\dfrac{du}{dx} * \dfrac{dy}{du}

Unparseable latex formula:

(2a - 1)(\dfrac{1}{6})u^-^0^.^5

Unparseable latex formula:

\dfrac{(2a - 1)}{6}(a^2 - a + 0.5)^-^0^.^5=

Unparseable latex formula:

\dfrac{dy}{dx}= \dfrac{(2a^3 - 3a^2 - 0.5)^-^0^.^5}{6}

Im not sure if this is right but if it is, can you just leave it in this form?:

Unparseable latex formula:

\dfrac{dy}{dx}= \dfrac{(2a - 1)(a^2 - a + 0.5)^-^0^.^5}{6}

OP

samir12

Unparseable latex formula:

\dfrac{(a^2-a+0.5)^0^.^5}{3}

Unparseable latex formula:

u = (a^2-a+0.5)}

\dfrac{du}{dx} = 2a - 1

Unparseable latex formula:

\dfrac{dy}{du} = (\dfrac{1}{2})(\dfrac{1}{3})u^-^0^.^5

\dfrac{du}{dx} * \dfrac{dy}{du}

Unparseable latex formula:

(2a - 1)(\dfrac{1}{6})u^-^0^.^5

Unparseable latex formula:

\dfrac{(2a - 1)}{6}(a^2 - a + 0.5)^-^0^.^5=

Unparseable latex formula:

\dfrac{dy}{dx}= \dfrac{(2a^3 - 3a^2 - 0.5)^-^0^.^5}{6}

Correct me if I'm wrong but I don't think you can expand the bracket in the last step...?

Hippysnake

Should I get

(2a-1)/(6(a^2-a+0.5)^0.5)

??

I assume thats a typo, so yes :smile:

:smile:

Well

Unparseable latex formula:

\dfrac{\mbox{d}}{\mbox{d}x} \left[ \dfrac{\left(a^2 - a + \frac{1}{2} \right)^{1/2}}{3} \right] = 0

, if that counts for anything. If you're differentiating with respect to

a

, on the other hand, then the chain rule works. Use

y = \dfrac{u^{1/2}}{3},\ u = a^2 - a + \dfrac{1}{2}

and it's fairly straightforward.

Hippysnake

Correct me if I'm wrong but I don't think you can expand the bracket in the last step...?

You're right; it should have been

\dfrac{2a - 1}{6\sqrt{a^2 - a + \frac{1}{2}}}

Hippysnake

Correct me if I'm wrong but I don't think you can expand the bracket in the last step...?

You can't.

OP

EierVonSatan

I assume thats a typo, so yes :smile:

:smile:

It's not a typo...I don't think so anyway

Eitherway, now I'm stuck on this step-

-5a^(-6) + (^that equation) = 0

Looks mighty... :s-smilie:

:s-smilie:

I am sorry the typing is so big. I would prefer it were smaller.

Hippysnake

It's not a typo...I don't think so anyway

So you did mean

\frac{2a-1}{6\sqrt{a^2-1+0.5}}

?

:s-smilie:

OP

EierVonSatan

So you did mean

\frac{2a-1}{6\sqrt{a^2-1+0.5}}

?

:s-smilie:

no.....?

EierVonSatan

So you did mean

\frac{2a-1}{6\sqrt{a^2-1+0.5}}

?

:s-smilie:

Why would there be a 1 there? There shouldn't be (and even if there was, -1 + 0.5 = -0.5).

OP

anyway guys, could you help with the next part?

-5a^(-6) + ^^that equation = 0?

Hippysnake

Should I get

(2a-1)/(6(a^2-1+0.5)^0.5)

??

EierVonSatan

Hippysnake

Should I get

(2a-1)/(6(a^2-a+0.5)^0.5)

??

I assume thats a typo, so yes :smile:

:smile:

Hippysnake

It's not a typo...I don't think so anyway

So it was a typo :smile:

:smile:

Hippysnake

anyway guys, could you help with the next part?

-5a^(-6) + ^^that equation = 0?

Which equation? The original one, or the differentiated one? Either way it seems like a fairly ridiculous equation to have to solve. You'd have to move the

\dfrac{-5}{a^6}

to the other side, square both sides and multiply through by

a^{12}

to start with... and that would be near impossible to solve without iteration or Wolfram|Alpha.

OP

nuodai

Which equation? The original one, or the differentiated one? Either way it seems like a fairly ridiculous equation to have to solve. You'd have to move the

\dfrac{-5}{a^6}

to the other side, square both sides and multiply through by

a^{12}

to start with... and that would be near impossible to solve without iteration or Wolfram|Alpha.

How about I scan in the question I'm doing?

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