Question

Differentiate implicitly to find d^2y/dx^2

x^3-3y^3=4

Answer 1

1) Differentiate both sides w.r.t. x. What do you get?

Answer 2

3x^2 and not sure what wrtx means :P

Answer 3

with relations to x?

Answer 4

w.r.t = with respect to.
Actually, you need to differentiate BOTH sides.
\[\frac{d}{dx}(x^3-3y^3)=\frac{d}{dx}(4)\]

Answer 5

Consider the left side:
\[\frac{d}{dx}(x^3-3y^3)=\frac{d}{dx}(x^3) - \frac{d}{dx}3y^3 = ...?\]

Answer 6

One point to keep in mind is that y is a function of x. So you need to differentiate it as such (as opposed to viewing it as a constant and dropping it from the derivative)

Answer 7

It's not just simply moving everything so y is by itself?

Answer 8

such as 4-x^3=-3y^3 dividing 3, getting the cube root then finding the derivative of that?

Answer 9

If you make y as the subject, I guess it is no longer "differentiate implicitly"

Answer 10

that would be an explicit differentiation. Sometimes functions cannot be expressed explicitly (thus they are also called implicit functions)

Answer 11

so in an implicit differentiation you apply the chain rule to f(y(x)), such as in this case: (3y^3)'=(9y^2)y'

Answer 12

Makes sense, so d^2y/dx^2 is just implicit differentiation? Not double implicit?

Answer 13

you will need to differentiate twice: both steps are implicit differentiation

Answer 14

then you identify the term d^2y/dx^2 and solve for it (which is what they ask for)

Answer 15

So what is the final answer? Taking my final tomorrow and teacher never really went over this, want to quickly brush up on it

Answer 16

2x/3y^2?

Answer 17

This is what i am getting: \[d^2/dx^2(x^3-3 y(x)^3 = 4\]\[6 x - 9 y(x) (y(x) y''(x)+2 y'(x)^2)=0\] Now you have to reshuffle things to solve for y''(x)

Answer 18

The first derivative is\[y'(x) = x^2/(3 y(x)^2)\] (note I am using y(x) to emphasize y is a function of x, i hope this is not confusing)

Answer 19

For example, here is how you would differentiate x^2+y^2=1 implicitly:
\[x^2+y^2=1\]\[\frac d{dx}(x+y)=\frac d{dx}1\]\[2x+\frac d{dx}y^2=0\]\[2x+\frac d{dy}y^2\frac{dy}{dx}=0\hspace{20pt}\mbox{(Chain rule)}\]\[2x+2y\frac{dy}{dx}=0\]\[\frac{dy}{dx}=-\frac yx\]

Answer 20

When you meet a function of y, you used the chain rule:
1. Differentiate the function wrt y
2. Multiply the derivative by \(\dfrac{dy}{dx}\)

Answer 21

@kc_kennylau The last step of your example should read \(\large\dfrac{dy}{dx} = -\dfrac{x}{y}\). ;)

Answer 22

oh yes sorry

Answer 23

Question about this problem!
http://gyazo.com/e2d41376f33f06aae1103cd8748fc7a4
Essentially it's just f'(x)=(3x^4+6)^3 when x=-1?

Answer 24

It's not f'(x) that is (3x^4+6)^3, it's f(x).

(However what we're finding is f'(x))

Answer 25

I know :O But after substituting in 3x^4+6 for u, it's what I said?

Answer 26

yep