Trying to solve for the derivative

[Mootjeuh]

Homework Statement

Hi, I'm trying to find the derivative of a specific equation and I feel like I messed up somewhere but can't figure out exactly what.

Homework Equations

$y = \frac{x^8}{8(1-x^2)^4}$

The Attempt at a Solution

First I calculated the derivatives for the denominator and numerator:
$\frac{\mathrm d}{\mathrm d x} \big( x^8 \big) = 8x^7$

$\frac{\mathrm d}{\mathrm d x} \big( 8(1-x^2)^4 \big) = 32(1-x^2)^3(-2x)$

Then the actual attempt at solving:
$\frac{\mathrm d}{\mathrm d x} \big( \frac{x^8}{8(1-x^2)^4} \big) = \frac{8(1-x^2)^48x^7-x^832(1-x^2)^3(-2x)}{[8(1-x^2)^4]^2}$

$\frac{8(1-x^8)8x^7-x^832(1-x^6)(-2x)}{[8(1-x^8)]^2} = \frac{8(1-x^8)8x^7-x^832(1-x^6)(-2x)}{(8-8x^8)^2 = 64-128x^8+64x^{16}}$

$\frac{8(1-x^8)8x^7-x^832(1-x^6)(-2x)}{64-128x^8+64x^{16}}$

 

[imiuru]

$(1-x^2)^4$ is not equal to $1-x^8$

 

[Mark44]

Homework Statement

Hi, I'm trying to find the derivative of a specific equation and I feel like I messed up somewhere but can't figure out exactly what.

Homework Equations

$y = \frac{x^8}{8(1-x^2)^4}$

The Attempt at a Solution

First I calculated the derivatives for the denominator and numerator:
$\frac{\mathrm d}{\mathrm d x} \big( x^8 \big) = 8x^7$

$\frac{\mathrm d}{\mathrm d x} \big( 8(1-x^2)^4 \big) = 32(1-x^2)^3(-2x)$

Then the actual attempt at solving:
$\frac{\mathrm d}{\mathrm d x} \big( \frac{x^8}{8(1-x^2)^4} \big) = \frac{8(1-x^2)^48x^7-x^832(1-x^2)^3(-2x)}{[8(1-x^2)^4]^2}$

$\frac{8(1-x^8)8x^7-x^832(1-x^6)(-2x)}{[8(1-x^8)]^2} = \frac{8(1-x^8)8x^7-x^832(1-x^6)(-2x)}{(8-8x^8)^2 = 64-128x^8+64x^{16}}$

$\frac{8(1-x^8)8x^7-x^832(1-x^6)(-2x)}{64-128x^8+64x^{16}}$

$(1-x^2)^4$ is not equal to $1-x^8$

Nor is (1 - x²)³ = 1 - x⁶

Your work to here looks fine:
$$\frac{8(1-x^2)^48x^7-x^832(1-x^2)^3(-2x)}{[8(1-x^2)^4]^2}$$
Expanding the (1 - x²) factors as you did is not helpful, even if you had done them correctly. Instead, find the greatest common factor of the two terms in the numerator, and factor it out of both terms. Notice that you have (1 - x²) to some power in both terms, and you have x to some power in both terms. There is also a coefficient of 64 in the first term and one of -64 in the second term.

 

[Mark44]

Mootjeuh,
Based on the mistakes in this thread and in the other thread you posted about logs, since you are apparently studying calculus, it would be very useful for you to spend some time reviewing algebra concepts from your previous classes. If you don't have a good working knowledge of how to manipulate algebra expressions, you will have a very difficult time following explanations in your calculus class.