# Nasty derivative

• February 8th 2010, 10:08 AM
wolfhound
Nasty derivative
Hello,
I was wondering if someone could show me how to do this?
Differentiate $a^{2} (a- a^{4} )^{3}$ Value of $a=x^{-2}$
I can't figure it out
• February 8th 2010, 10:15 AM
HallsofIvy
Quote:

Originally Posted by wolfhound
Hello,
I was wondering if someone could show me how to do this?
Differentiate $a^{2} (a- a^{4} )^{3}$ $a=x^{-2}$
I can't figure it out

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 $(x^n)'= nx^{n-1}$, you will get $0= -2x^{-3}$. 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, $(2a)(a- a^4)^3+ 3(a^2)(a- a^4)^2(1- 4a^3)= 0$.

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

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

I am not sure either
• February 8th 2010, 11:17 AM
vince
Quote:

Originally Posted by wolfhound
Hello,
I was wondering if someone could show me how to do this?
Differentiate $a^{2} (a- a^{4} )^{3}$ $a=x^{-2}$
I can't figure it out

$a^{3} (a- a^{4} )^{3} = \frac{1}{x^2}$ (lumped the $a$ with $a^2)$

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

$\frac{d}{dx} \times (a^3 \times (a-a^4)^3)$
=
$3a^2(a-a^4)^3\frac{da}{dx} + 3a^3(a-a^4)^{2} (1-4a^3)\frac{da}{dx}$
=
$3a^2(a-a^4)^2(a-a^4+a(1-4a^3))\frac{da}{dx}$
=
$3a^2(a-a^4)^2(2a-5a^4)\frac{da}{dx}$

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

$\Rightarrow$

$\frac{da}{dx} = \frac{-2}{(3a^2(a-a^4)^2(2a-5a^4)x^3}$
• February 8th 2010, 11:26 AM
wolfhound
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
• February 8th 2010, 11:38 AM
vince
just great...now u tell us. oh well, my work above was consistent with the ambiguity of your intial post.
• February 8th 2010, 11:48 AM
wolfhound
Sorry I thought it made sense
• February 8th 2010, 12:08 PM
vince
way too many little steps for typesetting through latex all the way through. at any rate, here is the setup:

you want

$\frac{d}{dx}(\frac{1}{x^4}(\frac{1}{x^2}-\frac{1}{x^8})^3)$ =
$\frac{d}{dx}( \frac{1}{x^4}(\frac{x^6-1}{x^8})^3)$
=
$\frac{d}{dx}( \frac{(x^6-1)^3}{x^{28}}$

Differentiating that is easy but lengthy.
• February 8th 2010, 12:41 PM
wolfhound
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?
• February 8th 2010, 01:11 PM
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

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

Was it your idea to substitute $a=x^{-2}$
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.
• February 9th 2010, 08:17 AM
wolfhound
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?
Thanks
• February 9th 2010, 01:08 PM
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...

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

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

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

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

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

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

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

Putting all these together

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

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

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

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

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

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

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

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

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

Therefore the derivative is

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

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

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

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

$=\frac{-10x^{18}+48x^{12}-66x^{6}+28}{x^{29}}$
• February 10th 2010, 06:06 AM
wolfhound
Thanks Archie,
you really are my saviour !(Clapping)