Question

1)
\[y=\frac{ (2x+1)^3 }{ 3x^2-1)^2 }, then find (dy/dx)
2)
Find
\[f'' (x) if f(x) =4(x^2+16)^{2/3}\]

Answer 1

there are 2 questions

Answer 2

for 1 use quotient rule and for 2 chain rule

Answer 3

your latex code broke

Answer 4

well the second suppose say:
find f'' if
\[f(x)=4(x^2+16)^{2/3}\]

Answer 5

second derivative... that's what we need. We need the combination of the chain rule and product rule

Answer 6

i'm not sure how to do it

Answer 7

I'm tired........ but wow.. ok product rule chain rule time.. chain rule is something like put the exponent in the front, subtract 1 from the exponent, take the derivative from the inside () and put it outside.

Answer 8

so here's an example for chain rule (x^2+5)^2
exponent goes in the front - >2
subtract 1 from the exponent -> 2-1 =1
the derivative of x^2+5 ---> 2x so that goes outside,,

my solution is

2(x^2+5)(2x) = 4x(x^2+5)

Answer 9

okay. can you check if I'm doing this correctly?
y' = \[6(2x+1)^2 ((3x^2-1)^{-2}+(2x^2+x))\]

Answer 10

which problem is this? Welll.... hmmm I would've used the quotient rule on it

Answer 11

this is for the first question