Question

help in finding second derivatives ((again i'm sorry lol))

Answer 1

the question is find the derivative of y=1/3x^2+4

Answer 2

@genilson1986

Answer 3

well write it using index notation

\[y=(3x^2 + 4)^{-1}\]

now use the chain rule

Answer 4

that will give the 1st derivative

for the 2nd derivative you will need the quotient rule.

Answer 5

if you want to use the quotient rule in your work you need to let

u = 1

the denominator of the fraction.

so

\[\frac{dy}{dx} = \frac{-6x}{(3x^2 + 4)^2}\]

now apply the quotient rule to this to find the 2nd derivative

Answer 6

ok so i redid it and used the chain rule and got -6x(3x^2+4)^-2, is that right?

Answer 7

yes... now you need the quotient rule

\[u = -6x ~~~~~~v = (3x^2 + 4)^2\]

Answer 8

You have the first derivative. To get the second, use the quotient rule. \(\frac{d(u/v)}{dv} = \frac{v*du - u*dv}{v^2}\)

Answer 9

@campbell_st is it not (3x^2+4) to the power of -2 instead of 2?

Answer 10

nope the denominator of the fraction is

\[(3x^2 + 4)^2\]

so that is what v is equal to

Answer 11

No, v is in the denominator so the power is positive. And I meant to say: \[\frac{d(u/v)}{dx}=\frac{v∗du−u∗dv}{v^2}\]

Answer 12

when you got the 1st derivative, you should rewrite it as an algebraic fraction.

this makes it easier to identify u and v

Answer 13

What the heck?
\[\frac{d(u/v)}{dx}=\frac{v*du-u*dv}{v^2}\] I'll stop.

Answer 14

but did i not have to bring -1 in front of (3x^2+4)^-1 and take away 1 from the power of -1 ?

Answer 15

the chain rule says to multiply

so its

\[\frac{dy}{dx} = -6x \times (3x^2 + 4)^{-2} = \frac{-6x}{3x^2 + 4}\]

so this should be the way you write the 1st derivative.

now when starting to find the 2nd derivative... you can use the quotient rule where

\[u = -6x ~~~~~~~~~v = (3x^2 + 4)^2\]

Answer 16

an alternative... is to take the 1st derivative as you have it written in your post and apply the product rule

Answer 17

\[f '(x) = (3x ^{2} +4)^{-2} *(- 6x)\]
or
\[f'(x) = \frac{ -6x }{ (3x ^{2} +4)^{2} }\]
you can chose both ways

Answer 18

remember the basic rule for differentiation

if \[y = x^n ~~~~~~~~~~~~~~\frac{dy}{dx} = n \times x^{n - 1}\]

when you used the chain rule for the 1st derivative you have it in index form where n = -1.

Answer 19

if you wanted to use the quotient rule as you attempted... to find the fist derivative

\[u = 1~~~~~~~~~~~v = 3x^2 + 4\]

then

\[\frac{du}{dx} = 0 ~~~~~~~~~~~~~~~~~~\frac{dv}{dx} = 6x\]

applying the quotient rule

\[\frac{dy}{dx} = \frac{(3x^2 + 4) \times 0 - (1) \times (6x)}{(3x^2 + 4)^2}\]


if you simplify this you get

\[\frac{dy}{dx} = \frac{-6x}{(3x^2 + 4)^2}\]

Answer 20

what i got for my first derivative using the chain rule was the same as @genilson1986 which is??