something doesn't seem right

[shadowfox853]

Either I'm not catching on to something or my math teacher is being cruel

the problem is f(x)= -4x/(9-x^2)^3 find the derivative using chain rule.

I know I'm supposed to use the quotient rule, but when I simplify the bottom it comes out with huge numbers like -x^6+27^4-243x^2+729.
I really think i'm missing something

 

[BigGlenntheHeavy]

$\displaystyle f(x) \ = \ \frac{-4x}{(9-x^{2})^{3}}$

$\displaystyle D_x[f(x)] \ = \ \frac{(-4)(9-x^{2})^{3}+4x(3)(9-x^{2})^{2}(-2x)}{(9-x^{2})^{6}}$

$\displaystyle = \ \frac{(-4)(9-x^{2})-24x^{2}}{(9-x^{2})^{4}} \ = \ -\frac{20x^{2}+36}{(9-x^{2})^{4}}$

 

[shadowfox853]

ok thank you very much

 

[galactus]

If I may, you can also do this with the product rule by rewriting as $\displaystyle -4x(9-x^{2})^{-3}$

$\displaystyle -4x\overbrace{(-3)(9-x^{2})^{-4}}^{\text{derivative of outside}}\cdot\underbrace{(-2x)}_{\text{derivative of inside}}+(9-x^{2})^{-3}(-4)$

simplify:

$\displaystyle \frac{-4(5x^{2}+9)}{(x^{2}-9)^{4}}$

Just another way to look at it.

 

[soroban]

Hello, shadowfox853!

Sheesh! . . . Never multiply out like that!

$\displaystyle \text{Differentiate: }\;f(x) \;=\;\frac{-4x}{(9-x^2)^3}$

$\displaystyle \text{We have: }\;f(x) \;=\;-4\cdot\frac{x}{(9-x^2)^3}$ . . . . . Yes, take the coefficient "out front"!


\(\displaystyle \text{Then: }\;f'(x) \;=\;-4\left[\frac{(9-x^2)^3\cdot 1 - x\cdot3(9-x^2)^2(-2x)}{(9-x^2}^6}\right] \;=\;-4\left[\frac{(9-x^2)^3 + 6x^2(9-x^2)^2}{(9-x^2)^6}\right]\)

$\displaystyle \text{Factor: }\;f'(x) \;=\;-4\,(9-x^2)^2\left[\frac{(9-x^2) + 6x^2}{(9-x^2)^6}\right] \;=\;-4\cdot\frac{9+5x^2}{(9-x^2)^4}$