Calculus problem

• Feb 24th 2009, 06:25 AM
Azathoth
Calculus problem
I am not sure if I am on the right track with this problem and where do i go from here. Please help.

Problem: y=4x^2(7x^2+8)^(-2/3) find dy/dx. This is my solution so far:

-8x^2(14x/3(7x^2+8)^(5/3) + 8x/(7x^2+8)^(2/3)
-112x^3/3(7x^2+8)^(5/3) + 8x/(7x^2+8)^(2/3)
Where do I go from here?

Thank you in advance. Any help is greatly appreciated.

• Feb 24th 2009, 06:58 AM
earboth
Quote:

Originally Posted by Azathoth
I am not sure if I am on the right track with this problem and where do i go from here. Please help.

Problem: y=4x^2(7x^2+8)^(-2/3) find dy/dx. This is my solution so far:

-8x^2(14x/3(7x^2+8)^(5/3) + 8x/(7x^2+8)^(2/3)
-112x^3/3(7x^2+8)^(5/3) + 8x/(7x^2+8)^(2/3)
Where do I go from here?

Thank you in advance. Any help is greatly appreciated.

Use the product rule (quotient rule is possible here too):

$\displaystyle f(x)=4x^2 \cdot (7x^2+8)^{-\frac23}$

$\displaystyle f'(x)= (7x^2+8)^{-\frac23} \cdot 8x + 4x^2 \cdot \left(-\frac23\right) (7x^2+8)^{-\frac53} \cdot 14x$

Transform into a quotient, the denominator is $\displaystyle 3(7x^2+8)^{-\frac53}$

$\displaystyle f'(x)=\dfrac{3 \cdot 8x((7x^2+8) - 2 \cdot 4x^2 \cdot 14x}{3(7x^2+8)^{\frac53}} = \dfrac{8x \cdot (7x^2+24)}{3(7x^2+8)^{\frac53}}$
• Feb 24th 2009, 07:48 AM
Azathoth
So was I on the wrong track or was I just going about it in a longer version? So the answer is [IMG]file:///C:/Users/Michelle/AppData/Local/Temp/moz-screenshot.jpg[/IMG]8x(7x^2 +24) over 3(7x^2 +8)^(5/3)? If so, eventually would I have gotten to that answer on the path I was taking? Thank you for your help.
• Feb 24th 2009, 10:37 AM
earboth
Quote:

Originally Posted by Azathoth
...

Problem: y=4x^2(7x^2+8)^(-2/3) find dy/dx. This is my solution so far:

-8x^2(14x/3(7x^2+8)^(5/3) + 8x/(7x^2+8)^(2/3)
-112x^3/3(7x^2+8)^(5/3) + 8x/(7x^2+8)^(2/3)
...

Quote:

Originally Posted by Azathoth
So was I on the wrong track or was I just going about it in a longer version?

...

I've got exactly the same result as you did. I only simplified the result using a convenient common denominator. Nothing else.