# Thread: how do u take the second derivative wen it doesn't start with a "y="

1. ## how do u take the second derivative wen it doesn't start with a "y="

how do u find the second derivative of this

3y^2 - 4x = 16

2. Originally Posted by calchelp
how do u find the second derivative of this

3y^2 - 4x = 16
we can solve for y first (which isn't hard) and then take the derivative twice, or we can take the derivative implicitly, which would you prefer? Have you learnt implicit differentiation?

3. can u briefly explain implicit diffrentiation cause thats only one from the many that i have...and the entire section is implicit diffrentiation

4. Originally Posted by calchelp
can u briefly explain implicit diffrentiation cause thats only one from the many that i have...and the entire section is implicit diffrentiation
In implicit differentiation, we take nothing for granted. we always identify what we are taking the derivative of, and with respect to what. That is, if we differentiate a y-term with respect to x, we attach the notation $\frac {dy}{dx}$ to it. This means we took the derivative of y with respect to x. We can simply call this y'. We don't attach anything if we differentiate a variable that is the same as the variable that we are differentiating with respect to. That is, if we differentiate x with respect to x, we don't attach anything. Why is that? Technically we do, if we differentiate an x-term with respect to x, we attach $\frac {dx}{dx}$, but since derivative notations can function as fractions, we simplify this to 1, so you don't see it. If we differentiate a term with both x's and y's in it, we use the product rule. and we attach the appropriate notation for each part. when we differentiate the x-term we attach nothing. when we differentiate the y-term, we attach $\frac {dy}{dx}$

Let's see how this works

Originally Posted by calchelp
how do u find the second derivative of this

3y^2 - 4x = 16
$3y^2 - 4x = 16$

We proceed by Implicit differentiation

$\Rightarrow 6y~y' - 4 = 0$ .......remember, the derivative of a constant is zero, so the 16 disappears. we attach y' to the derivative of the y-term, but nothing to the derivative of the x-term as explained above.

Now we simply solve for y'

$\Rightarrow y' = \frac {4}{6y}$

Now try the second derivative

5. ## tell me if this is the right approach

so the first derivative is

y'=4/6y

y''=[6y * 0] - [4 * 6y'] /(6y^2)

y"= -4 * 6 y' / (36y^2)

what do we do with the 6y' left in the equation?

6. Originally Posted by calchelp
so the first derivative is

y'=4/6y

y''=[6y * 0] - [4 * 6y'] /(6y^2)

y"= -4 * 6 y' / (36y^2)

what do we do with the 6y' left in the equation?
you can replace it with the original expression for y' if it bothers you, so you could replace it with 4/6y