Question

Can anyone help me on this problem regarding implicit differentiation? Thanks
http://prntscr.com/7u700y

Answer 1

when differentiating y, remember to use the chain rule...

Answer 2

For example,

\[\frac{ d }{ dx }[y^3] = \frac{ d }{ dy }[y^3]*\frac{ dy }{ dx }\]

Answer 3

Basic chain rule for this

\[\frac{ d }{ dx } = \frac{ d }{ dy }*\frac{ dy }{ dx }\]

Answer 4

Following the above example,

\[\frac{ d }{ dx }[y^3] = 3y^2*\frac{ dy }{ dx }\]

Answer 5

Do that for the y terms, and differentiate the x terms normally, then solve for dy/dx

Answer 6

Okay I'll try it and tell you what I get.

Answer 7

k

Answer 8

Okay I am actually still very confused. Am i supposed to find the second derivative of
5x^2 + y4 = −9 and then use implicit differentiation?

Answer 9

Yes you want to find the second derivative of the given thing with respect to X, but to take the derivative w.r.t. X of the Y term, you have to use that chain rule... here ill type it out

Answer 10

\[\frac{ d }{ dx }[5x^2]+\frac{ d }{ dx }y^4 = \frac{ d }{ dx }[-9]\]

\[\frac{ d }{ dx }[5x^2]+\frac{ d }{ dy }y^4*\frac{ dy }{ dx } = \frac{ d }{ dx }[9]\]

Answer 11

now take derivative of y term with respect to y like normal, but you have to put the y' on there too from the chain rule.

Answer 12

\[\frac{ d }{ dx }[5x^2]+\frac{ d }{ dy }y^4*\frac{ dy }{ dx } = \frac{ d }{ dx }[-9]\]

\[10x^2+4y^3*\frac{ dy }{ dx } = 0\]

Answer 13

Solve that for dy/dx and that is your first derivative...

Answer 14

Or just remember, if you are taking the derivative of a Y term, tag on a dy/dx next to it ...basically

Answer 15

Okay for the first derivative I got : dy/dx= -10x^2 / 4y^3

Answer 16

And then take the derivative again?

Answer 17

So you get,

\[\frac{ dy }{ dx } = \frac{ -10x^2 }{ 4y^3 }\]

right

Answer 18

This time you need the product or the quotient rule and the chain rule...

Answer 19

I can start typing it out...

Answer 20

Im using the quotient rule

Answer 21

\[\frac{ d }{ dx }\frac{ dy }{ dx } = \frac{ d^2y }{ dx^2 }\]

k, ill start typing it

Answer 22

I got : -80xy^3 +120x^2y^2 / (4y^3)^2

Answer 23

\[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx}\frac{ -5x^2 }{ 2y^3 }\]

Answer 24

Oh I think I took the derivative of the wrong thing?

Answer 25

So it would be -20 when you plug in x and y

Answer 26

\[\frac{ d^2y }{ dx^2 }=\frac{ 2y^2*-10x - [-5x*6y^2*\frac{ dy }{ dx } ]}{ 4y^6 }\]

Answer 27

That is the quotient rule , using the chain rule on the y term in the numorator.

Answer 28

So am I supposed to plug in x and y into which equation?

Answer 29

yes, but notice the second derivative has a term of the first derivative in the numerator... but you know what dy/dx is , so sub that in first

Answer 30

\[\frac{ dy }{ dx } = \frac{ -5x^2 }{ 2y^3 }\]

reduced fraction...
Sub that in for dyy/dx in the second derivative
then plug in the x and y and solve...

Answer 31

Less writing overall if you want to use y' and y'' instead of dy/dx things

Answer 32

goodluck, i dont want to calculate it.

Answer 33

\[\large \color{blue}{\checkmark}\]

Answer 34

I got -310

Answer 35

Thanks for all the help!

Answer 36

yw, just get that Leibanitz notation dy/dx crap down, it makes the chain rule and this stuff easier i think .. goodluck

Answer 37

Okay I will try using it more.