# Homework Help: Find The Derivative

1. Feb 27, 2010

### polak333

1. The problem statement, all variables and given/known data
=$$\frac{-3x^{4}}{(4x-8)^{1/2}}$$

Is it actually correct, I'm not sure if it's correct, still.

2. Relevant equations

Quotient Rule and Chain Rule

3. The attempt at a solution
=$$\frac{-3x^{4}}{(4x-8)^{1/2}}$$

=$$\frac{(-12x^{3})(4x-8)^{1/2}-(-3x^{4})(1/2)(4x-8)^{-1/2}(4)}{[(4x-8)^{1/2}]^{2}}$$

=$$\frac{-12x^{3}(4x-8)^{1/2}+6x^{4}(4x-8)^{-1/2}}{(4x-8)}$$

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

I'm not sure if it's correct up to here, but the $$(4x-8)^{1/2}$$ isn't working. If there was no $$^{1/2}$$ it would work something like this:

=$$\frac{-6x^{3}(8x-16-x)}{(4x-8)^{3/2}}$$

=$$\frac{-6x^{3}(7x-16)}{(4x-8)^{3/2}}$$

But still unsure how they get the -3 in front and not -6 like I got.

=$$\frac{-3x^{3}(7x-16)}{(4x-8)^{3/2}}$$

Any help is appreciated. Thank you!

Look 2 posts lower for CLEARED UP version!

Still looking for help!

Last edited: Feb 27, 2010
2. Feb 27, 2010

### VeeEight

In '3. The attempt at a solution' - I am not sure how you went from the second last to last equation (before you say "I'm not sure if it's correct up to here"). I have not gone through the whole thing.

3. Feb 27, 2010

### snipez90

To the OP, I think your answer is correct, whereas whatever the answer at the very end is in fact incorrect. Another way to see this is to multiply the top and bottom of
$$\frac{-12x^{3}(4x-8)^{1/2}+6x^{4}(4x-8)^{-1/2}}{(4x-8)}$$
by (4x-8)^(1/2). The numerator of the resulting expression simplifies to -12x^3(4x-8) + 6x^4 which after more algebra comes down to your expression.

4. Feb 27, 2010

### polak333

Ok, starting from 3 again:

All I'm using is the Quotient Rule here.

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

=$$\frac{(-12x^{3})(4x-8)^{1/2} - (-3x^{4})(1/2)(4x-8)^{-1/2}(4)}{[(4x-8)^{1/2}]^{2}}$$

=$$\frac{(-12x^{3})(4x-8)^{1/2}-(-3x^{4})(1/2)(4x-8)^{-1/2}(4)}{(4x-8)}$$

=$$\frac{-12x^{3}(4x-8)^{1/2} + 3x^{4}(1/2)(4)(4x-8)^{-1/2}}{(4x-8)}$$

=$$\frac{-12x^{3}(4x-8)^{1/2} + 3x^{4}(2)(4x-8)^{-1/2}}{(4x-8)}$$

=$$\frac{-12x^{3}(4x-8)^{1/2} + 6x^{4}(4x-8)^{-1/2}}{(4x-8)}$$

Right here, I bring down the $$(4x-8)^{-1/2}$$, to make it positive $$(4x-8)^{1/2}$$

=$$\frac{-12x^{3}(4x-8)^{1/2} + 6x^{4}}{(4x-8)(4x-8)^{1/2}}$$

=$$\frac{-12x^{3}(4x-8)^{1/2} + 6x^{4}}{(4x-8)^{3/2}}$$

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

Then the rest... (which is apparently wrong somewhere in the question)

=$$\frac{-6x^{3}(8x-16-x)}{(4x-8)^{3/2}}$$

=$$\frac{-6x^{3}(7x-16)}{(4x-8)^{3/2}}$$

=$$\frac{-3x^{3}(7x-16)}{(4x-8)^{3/2}}$$