Question

differentiate
[(x^2 + 9) / (4- x^2)] ^4

Answer 1

\[\Large \frac{d}{dx}\left[\frac{x^2+9}{x-x^2}\right]^4\quad=\quad 4\left[\frac{x^2+9}{x-x^2}\right]^3\cdot \color{royalblue}{\left[\frac{x^2+9}{x-x^2}\right]'}\]

Answer 2

So we start with the power rule, yes?
Then we chain rule, multiplying by the derivative of the inner function.

Answer 3

Looks like we'll need to apply the quotient rule for this part.

Answer 4

\[\Large =4\left[\frac{x^2+9}{x-x^2}\right]^3\cdot \left[\frac{\color{royalblue}{(x^2+9)'}(x-x^2)-(x^2+9)\color{royalblue}{(x-x^2)'}}{(x-x^2)^2}\right]\]

Answer 5

So there's our quotient rule setup.
Make sense? :o

Answer 6

i don't get the power rule.

Answer 7

The outermost function is:\[\Large (stuff)^4\]Remember your Power Rule for derivatives? :)\[\Large (x^n)' \quad=\quad n\cdot x^{n-1}\]

Answer 8

okay. got that.

Answer 9

\[4\left( \frac{ x^2+9 }{ x-x^2 }\right)^3 \left( \frac{ (1-2x)(x^2+9)-(x-x^2 )(2x) }{ (x-x^2)^2 } \right)\]

Answer 10

sorry a little change.

Answer 11

I'm sorry. I'm confused. Sorry to butt in. But isn't the denominator in the original problem \[4-x ^{2}\]?

Answer 12

yep. i changed it, but it was after the help. I have the power rule, but not sure if I set up the rest right?

Answer 13

With the chain rule for division, we learned this:

Lo dee hi minus hi dee lo. Strike a line and square below.

This means what's on the bottom (lo) times the derivative of what's on the top (hi) subtract what's on the top (hi) times the derivative of what's on the bottom (lo). Then draw your dividing line (some people have a name for it, but I don't) and square what's on the bottom.

Answer 14

got that

Answer 15

So, then you should have the answer. Depending on your teacher, you may have to simplify, but that's just algebra. I would guess you wouldn't.

Answer 16

thanks