Hello,

I was wondering if someone could show me how to do this?

Differentiate$\displaystyle a^{2} (a- a^{4} )^{3}$ Value of $\displaystyle a=x^{-2}$

I can't figure it out

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- Feb 8th 2010, 10:08 AMwolfhoundNasty derivative
Hello,

I was wondering if someone could show me how to do this?

Differentiate$\displaystyle a^{2} (a- a^{4} )^{3}$ Value of $\displaystyle a=x^{-2}$

I can't figure it out - Feb 8th 2010, 10:15 AMHallsofIvy
Do you mean "differentiate both sides of this equation with respect to x"? If so, since the left side is just a constant with respect to x, and $\displaystyle (x^n)'= nx^{n-1}$, you will get $\displaystyle 0= -2x^{-3}$. That equation is never true however.

If you mean "differentiate both sides of this equation with respect to a", then, since the right side is a constant with respect to a, and using the product rule and chain rule, $\displaystyle (2a)(a- a^4)^3+ 3(a^2)(a- a^4)^2(1- 4a^3)= 0$.

If neither of those is what you mean, then I have no idea what you**do**mean. - Feb 8th 2010, 10:47 AMmabruka
The other posible thing going on is that we wants to differentiate both sides like when you do do normal substitution in integrals:

$\displaystyle u=x^2 \Longrightarrow du=2xdx$

I am not sure either - Feb 8th 2010, 11:17 AMvince
$\displaystyle a^{3} (a- a^{4} )^{3} = \frac{1}{x^2} $ (lumped the $\displaystyle a$ with $\displaystyle a^2)$

Solve for $\displaystyle \frac{da}{dx}$ via implicit differentiation ($\displaystyle a$ is clearly a function of $\displaystyle x$)

$\displaystyle \frac{d}{dx} \times (a^3 \times (a-a^4)^3)$

=

$\displaystyle 3a^2(a-a^4)^3\frac{da}{dx} + 3a^3(a-a^4)^{2} (1-4a^3)\frac{da}{dx}$

=

$\displaystyle 3a^2(a-a^4)^2(a-a^4+a(1-4a^3))\frac{da}{dx}$

=

$\displaystyle 3a^2(a-a^4)^2(2a-5a^4)\frac{da}{dx}$

Also

$\displaystyle \frac{d}{dx}\frac{1}{x^2} = -2x^{-3}$

$\displaystyle \Rightarrow$

$\displaystyle \frac{da}{dx} = \frac{-2}{(3a^2(a-a^4)^2(2a-5a^4)x^3}$ - Feb 8th 2010, 11:26 AMwolfhound
Hello

The value of a= x^-2

yes I need to use the chain and product rule but all these exponents are mixing me up help,Please - Feb 8th 2010, 11:38 AMvince
just great...now u tell us. oh well, my work above was consistent with the ambiguity of your intial post.

- Feb 8th 2010, 11:48 AMwolfhound
Sorry I thought it made sense

Please show me - Feb 8th 2010, 12:08 PMvince
way too many little steps for typesetting through latex all the way through. at any rate, here is the setup:

you want

$\displaystyle \frac{d}{dx}(\frac{1}{x^4}(\frac{1}{x^2}-\frac{1}{x^8})^3)$ =

$\displaystyle \frac{d}{dx}( \frac{1}{x^4}(\frac{x^6-1}{x^8})^3)$

=

$\displaystyle \frac{d}{dx}( \frac{(x^6-1)^3}{x^{28}}$

Differentiating that is easy but lengthy. - Feb 8th 2010, 12:41 PMwolfhound
Hi why raise x^8 to the power of 3,and not raise (x^6-1) to power of 3 aslo?

is it easier to leave the top in brackets? - Feb 8th 2010, 01:11 PMArchie Meade
Hi wolfhound,

Vince has simply left the numerator "written" as a cube,

since you can now apply the chain rule to it,

or simply multiply out

$\displaystyle \left(x^6-1\right)\left(x^6-1\right)\left(x^6-1\right)$

Was it your idea to substitute $\displaystyle a=x^{-2}$

in the beginning?

Or was that requested by the question?

You see, you can do things the short way using x,

then manipulating the fraction to a simpler manageable one as Vince did.

Or... use the messy substitution and get some good experience with the Chain Rule.

Let me know the drive of the question,

I can show you the way using the substitution if you want to play along with it. - Feb 9th 2010, 08:17 AMwolfhound
Hello Archie (Hi)

It wasn't my Idea to use a=x^-2

It was in the question,

I cant seem to get this answer ( 28 -66x^6+48x^12-10x^18)/x^29

Is it best to use the product+chain rule, or chain+quotient rule?

Please help

Thanks - Feb 9th 2010, 01:08 PMArchie Meade
Hi wolfhound,

Since you have a product after your substitution,

you need to use a combination of product and chain rule.

It's much faster to do it the way Vince showed.

Nevertheless, here's what to do with the "a" substitution...

$\displaystyle \frac{d}{dx}\left(a^2\left(a-a^4\right)^3\right)=\frac{d}{dx}(uv)=v\frac{du}{dx }+u\frac{dv}{dx}=v\frac{da}{dx}\frac{du}{da}+u\fra c{da}{dx}\frac{dv}{da}$

$\displaystyle u=a^2\ \Rightarrow\ \frac{du}{da}=2a$

$\displaystyle v=\left(a-a^4\right)^3=w^3$

$\displaystyle \frac{dv}{dw}=3w^2$

$\displaystyle w=a-a^4\ \Rightarrow\ \frac{dw}{da}=1-4a^3$

$\displaystyle \frac{dv}{da}=\frac{dv}{dw}\frac{dw}{da}=3w^2\left (1-4a^3\right)=3\left(a-a^4\right)^2\left(1-4a^3\right)$

$\displaystyle a=x^{-2}\ \Rightarrow\ \frac{da}{dx}=-2x^{-3}$

Putting all these together

$\displaystyle \frac{d}{dx}\left(a^2\left(a-a^4\right)^3\right)=v\frac{da}{dx}\frac{du}{da}+u\ frac{da}{dx}\frac{dv}{dw}\frac{dw}{da}$

We can write all of these terms using x to get the answer you are seeking.

$\displaystyle u=a^2=x^{-4}$

$\displaystyle v=\left(a-a^4\right)^3=\left(x^{-2}-x^{-2(4)}\right)^3=\left(x^{-2}-x^{-8}\right)^3=\left(x^{-2}-x^{-8}\right)$$\displaystyle \left(x^{-2}-x^{-8}\right)\left(x^{-2}-x^{-8}\right)$

$\displaystyle =\left(x^{-4}-2x^{-10}+x^{-16}\right)\left(x^{-2}-x^{-8}\right)=x^{-6}-3x^{-12}+3x^{-18}-x^{-24}$

$\displaystyle \frac{da}{dx}=-2x^{-3}$

$\displaystyle \frac{du}{da}=2x^{-2}$

$\displaystyle \frac{dv}{dw}=3w^2=3\left(a-a^4\right)^2=3\left(a^2-2a^5+a^8\right)=3\left(x^{-4}-2x^{-10}+x^{-16}\right)$

$\displaystyle \frac{dw}{da}=1-4a^3=1-4x^{-6}$

Therefore the derivative is

$\displaystyle \left(x^{-6}-3x^{-12}+3x^{-18}-x^{-24}\right)\left(-2x^{-3}\right)2x^{-2}+x^{-4}\left(-2x^{-3}\right)$$\displaystyle 3\left(x^{-4}-2x^{-10}+x^{-16}\right)\left(1-4x^{-6}\right)$

$\displaystyle =-4x^{-11}+12x^{-17}-12x^{-23}+4x^{-29}-6x^{-7}$$\displaystyle \left(x^{-4}-4x^{-10}-2x^{-10}+8x^{-16}+x^{-16}-4x^{-22}\right)$

$\displaystyle =4x^{-11}+12x^{-17}-12x^{-23}+4x^{-20}-6x^{-11}+36x^{-17}-54x^{-23}+24x^{-29}$

$\displaystyle =-10x^{-11}+48x^{-17}-66x^{-23}+28x^{-29}$

$\displaystyle =\frac{-10x^{18}+48x^{12}-66x^{6}+28}{x^{29}}$ - Feb 10th 2010, 06:06 AMwolfhound
Thanks Archie,

you really are my saviour !(Clapping)