Question

find dy/dx (3x^2+2/1-x^3)^5 using chain rule or derivatives if a qoutient

Answer 1

\[\large y=\left(\cfrac{3x^2 +2}{1-x^3}\right)^5\]\[\large y' = 5\left(\cfrac{3x^2+2}{1-x^3}\right)^4 \cdot \cfrac{d}{dx} \left(\cfrac{3x^2+2}{1-x^3}\right)\]

Answer 2

Thought i'd just write out the derivative separately.

Answer 3

May I interrupt for a moment?

Answer 4

hm?

Answer 5

\[\large \text{quotient rule} = \cfrac{f'g -g'f}{g^2}\]\[\large f' = 6x\]\[\large g' = -3x^2\] Can you apply the quotient rule to the derivative?

Answer 6

yah...

Answer 7

Can you write it or draw it out and tell me what you get? :)

Answer 8

Well I already did, so I'll continue. Fun fact, you never need to memorize the quotient rule if you know your algebra. I'd always get my signs confused on the numerator, so eventually I gave up trying to memorize it, and started solving everything this way:
\[(\frac{f(x)}{g(x)})'=(f(x)*(g(x))^{-1})=f'(x)*(g(x))^{-1}+(-1)(g(x))^{-2}*g'(x)*f(x)\]

Answer 9

Yeah, that's another way.