Find Derivative of \frac{-3x^4}{(4x-8)^{1/2}}

[polak333]

Homework Statement

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

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

Homework Equations

Quotient Rule and Chain Rule

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.

Answer:
=\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!

 

[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.

 

[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
<br /> \frac{-12x^{3}(4x-8)^{1/2}+6x^{4}(4x-8)^{-1/2}}{(4x-8)}<br />
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.

 

[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}}

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

I hope this cleared it up a lot.