1. ## Nasty 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

2. Originally Posted by wolfhound
Hello,
I was wondering if someone could show me how to do this?
Differentiate$\displaystyle a^{2} (a- a^{4} )^{3}$ $\displaystyle a=x^{-2}$
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 $\displaystyle (x^n)'= nx^{n-1}$, 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.

3. 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

4. Originally Posted by wolfhound
Hello,
I was wondering if someone could show me how to do this?
Differentiate$\displaystyle a^{2} (a- a^{4} )^{3}$ $\displaystyle a=x^{-2}$
I can't figure it out
$\displaystyle a^{3} (a- a^{4} )^{3} = \frac{1}{x^2}$ (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
$\displaystyle \frac{d}{dx}\frac{1}{x^2} = -2x^{-3}$

$\displaystyle \Rightarrow$

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

5. 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

6. just great...now u tell us. oh well, my work above was consistent with the ambiguity of your intial post.

7. Sorry I thought it made sense

8. 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.

9. 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?

10. 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.

11. Hello Archie
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

12. 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)$

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

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

$\displaystyle =\frac{-10x^{18}+48x^{12}-66x^{6}+28}{x^{29}}$

13. Thanks Archie,
you really are my saviour !